intermediate (toss) course senior secondary course

SCERT Campus, Opp. to L.B. Stadium, Basheerbagh, Hyderabad - 500 001, Telangana.Phone : 040-23299568, website : telanganaopenschool.org. E-mail : [email protected]

Telangana Open School Society (TOSS), Hyderabad

Intermediate (TOSS) CourseSenior Secondary Course

MATHEMATICS

311311311311311

(CORE MODULES)

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Designed and developed

Suvendu Sekhar Das

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© Telangana Open School Society

No of Copies : 1153

Published by the Director Telangana Open School Society, Hyderabad with theCopyright Permission of NIOS, NOIDA.

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Reprint : 2018

FORE WARD

Education plays an important role in the modern society. Many innovations can beachieved through Education. Hence the Department of Education is giving equalimportance to non-formal education through Open Distance Learning (ODL) mode onthe lines of formal education. This is the first State Open School established in thecountry in the year 1991 offering courses up to Upper primary Level till 2008. Fromthe academic year 2008-2009 SSC Course was introduced and Intermediate Coursefrom the year 2010-2011. The qualified learners from the Open School are eligible forboth higher studies and employment. So far 7,67,190 learners were enrolled in theOpen Schools and 4,50,024 learners have successfully completed their courses.

With the aim of improving the administration at the grass-root level the TelanganaGovernment re-organized the existing Districts and formed 31 new Districts. Theformation of new Districts provide wide range of employment and Businessopportunities besides self employment. Given the freedom and flexibilities available,the Open School system provides a second chance of learning for those who could notfulfill their dreams of formal education.

Government of Telangana is keen in providing quality education by supplying studymaterials along with the text books to enable the learners to take the exam with ease.Highly experienced professionals and subject experts are involved in preparingcurriculum and study material based on subject wise blue prints. The study material forthe academic year 2018-19 is being printed and supplied to all the learners throughoutthe state.

I wish the learners of Open School make best use of the study material to brightentheir future opportunities and rise up to the occasion in building Bangaru Telangana.

With best wishes ........

S. Venkateshwara SharmaDIRECTOR,

Telangana Open School Society,Hyderabad

MATHEMATICS 1

Notes

MODULE - IVFunctions

Sets, Relations and Functions

15

SETS, RELATIONS AND FUNCTIONS

Let us consider the following situation :

One day Mrs. and Mr. Mehta went to the market. Mr. Mehta purchased the following objects/items.

"a toy, one kg sweets and a magazine". Where as Mrs. Mehta purchased the following objects/items.

"Lady fingers, Potatoes and Tomatoes".

In both the examples, objects in each collection are well defined. What can you say about thecollection of students who speak the truth ? Is it well defined? Perhaps not.

A set is a collection of well defined objects. For a collection to be a set it is necessary that itshould be well defined.

The word well defined was used by the German Mathematician George Cantor (1845- 1918A.D) to define a set. He is known as father of set theory. Now-a-days set theory has becomebasic to most of the concepts in Mathematics.

In our everyday life we come across different types of relations between the objects. Theconcept of relation has been developed in mathematical form.

The word function was introduced by Leibnitz in 1694. Function is a special type of relation.Each function is a relation but each relation is not a function. In this lesson we shall discusssome basic definitions and operations involving sets, Cartesian product of two sets, relationbetween two sets, the conditions under when a relation becomes a function, different types offunction and their properties.

MATHEMATICS

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After studying this lesson, you will be able to :

� define a set and represent the same in different forms;

� define different types of sets such as, finite and infinite sets, empty set, singleton set,equivalent sets, equal sets, sub sets and cite examples thereof;

� define and cite examples of universal set, complement of a set and difference betweentwo sets;

� define union and intersection of two sets;

� represent union and intersection of two sets, universal sets, complement of a set, differencebetween two sets by Venn Diagram;

� solve real life problems using Venn Diagram;

� define Cartesian product of two sets;

� define relation, function and cite examples thereof;

� find domain and range of a function;

� define and cite examples of diferent types of functions (one-one, many-one, onto, intoand bijection);

� determine wheather a function is one-one, many-one, onto or into;

� draw the graph of functions;

� define and cite examples of odd and even functions;

� determine wheather a function is odd or even or neither;

� define and cite examples of functions like | x |, [x] the greatest integer function, polynomialfunctions, logarithmic and exponential functions;

� define composition of two functions;

� define the inverse of a function; and

� state the conditions for the inverse to exist.

EXPECTED BACKGROUND KNOWLEDGE� Number systems, concept of ordered pairs.

OBJECTIVES

MATHEMATICS 3

Notes



15.1 SOME STANDARD NOTATIONSBefore defining different terms of this lesson let us consider the following examples:

(i) collection of tall students in your school.

(ii) collection of honest persons in yourcolony.

(iii) collection of interesting books in yourschool library.

(iv) collection of intelligent students in yourschool.

(i) collection of those students of your schoolwhose height is more than 180 cm.

(ii) collection of those people in your colonywho have never been found involved in anytheft case.

(iii) collection of Mathematics books in yourschool library.

(iv) collection of those students in your schoolwho have secured more than 80% marks inannual examination.

In all collections written on left hand side of the vertical line the term tallness, interesting, honesty,intelligence are not well defined. In fact these notions vary from individual to individual. Hencethose collections can not be considered as sets.While in all collections written on right hand side of the vertical line, 'height' 'more than 180 cm.''mathematics books' 'never been found involved in theft case,' ' marks more than 80%' are welldefined properties. Therefore, these collections can be considered as sets.If a collection is a set then each object of this collection is said to be an element of this set. A setis usually denoted by capital letters of English alphabet and its elements are denoted by smallletters.

For example, A = Toy elephant, packet of sweets, magazines.Some standard notations to represent sets :

N : the set of natural numbers

W : the set of whole numbers

Z or I : the set of integers

Z+ : the set of positve integers

Z : the set of negative integers

Q : the set of rational numbers

R : the set of real numbers

C : the set of complex numbersOther frequently used symbols are :

: 'belongs to'

: 'does not belong to'

: There exists, : There does not exist.

MATHEMATICS

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For example N is the set of natural numbers and we know that 2 is a natural number but 2 isnot a natural number. It can be written in the symbolic form as 2 N and 2 N .

15. 2 REPRESENTATION OF A SETThere are two methods to represent a set.15.2.1 (i) Roster method (Tabular form)In this method a set is represented by listing all its elements, separating these by commas andenclosing these in curly bracket.If V be the set of vowels of English alphabet, it can be written in Roster form as :

V = { a, e, i, o, u}(ii) If A be the set of natural numbers less than 7.then A={1, 2, 3, 4, 5, 6}, is in the Roster form.

Note : To write a set in Roster form elements are not to be repeated i.e. all elements aretaken as distinct. For example if A be the set of letters used in the word mathematics, then

A = {m, a, t, h, e, i, c, s}

15.2.2 Set-builder formIn this form elements of the set are not listed but these are represented by some commonproperty.Let V be the set of vowels of English alphabet then V can be written in the set builder form as:

V = {x : x is a vowel of English alphabet}(ii) Let A be the set of natural numbers less than 7.

then A = {x : x N and 1 x 7}

Note : Symbol ':' read as 'such that'

Example: 15.1: Write the following in set -builder form :

(a) A={ 3, 2, 1,0,1,2,3} (b) B = {3,6,9,12}

Solution : (a) A = {x : x Z and 3 x 3 }

(b) B = {x : x 3n and n N,n 4 }

Example 15.2 Write the following in Roster form.

(a) C = { x : x N and 50 x 60 }

(b) D = 2x : x R and x 5x 6 0

Solution : (a) C ={50, 51, 52,53,54,55,56,57,58,59,60}

(b) 2x 5x 6 0

x 3 x 2 0

MATHEMATICS 5

Notes



x 3,2 .

D 2,3

15.3 CLASSIFICATION OF SETS15.3.1 Finite and infinite sets

Let A and B be two sets whereA = {x : x is a natural number}B = {x : x is a student of your school}

As it is clear that the number of elements in set A is not finite (infinite) while number of elements inset B is finite. A is said to be an infinite set and B is said to be is finite set.

A set is said to be finite if its elements can be counted and it is said to be infiniteif it is not possible to count upto its last element.

15.3.2 Empty (Null) Set : Consider the following sets.

A = 2x:x R and x 1 0

B = {x : x is number which is greater than 7 and less than 5}Set A consists of real numbers but there is no real number whose square is 1. Therefore thisset consists of no element. Similiarly there is no such number which is less than 5 and greaterthan 7. Such a set is said to be a null (empty) set. It is denoted by the symbol void, or { }

A set which has no element is said to be a null/empty/void set, and is denoted by .

15.3.3 Singleton Set : Consider the following set :A = {x : x is an even prime number}

As there is only one even prime number namely 2, so set A will have only one element. Such a

set is said to be singleton. Here A = 2 .

A set which has only one element is known as singleton.15.3.4 Equal and equivalent sets : Consider the following examples.

(i) A = 1,2,3 , B = 2,1,3 (ii) D = 1,2,3 , E = a,b,c .

In example (i) Sets A and B have the same elements. Such sets are said to be equal sets and itis written as A = B. In example (ii) set D and E have the same number of elements but elementsare different. Such sets are said to be equivalent sets and are written as A B.Two sets A and B are said to be equivalent sets if they have same number of elements but theyare said to be equal if they have not only the same number of elements but elements are also thesame.15.3.5 Disjoint Sets : Two sets are said to be disjoint if they do not have any common element.For example,sets A= { 1,3,5} and B = { 2,4,6 } are disjoint sets.

MATHEMATICS

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xample 15.3 Given that

A = { 2, 4} and B = { x : x is a solution of 2x 6x 8 0 }

Are A and B disjoint sets ?

Solution : If we solve 2x 6x 8 0 ,we get

x 4, 2 .

B 4, 2 and A 2,4

Clearly ,A and B are disjoint sets as they do not have any common element.

Example 15.4 If A = {x : x is a vowel of English alphabet}

and B = y : y N and y 5

Is (i) A = B (ii) A B ?Solution : A = {a, e, i, o, u}, b = {1, 2, 3, 4,5}.

Each set is having five elements but elements are differentA B but A B.

Example 15.5 Which of the following setsA = {x : x is a point on a line}B = {y : y N and y 50}are finite or infinite ?

Solution : As the number of points on a line is uncountable (cannot be counted) so A is aninfinite set while the number of natural numbers upto fifty can be counted so B is a finite set.

Example 15.6 Which of the following sets

A = 2x :x i s irrational and x 1 0 .

B = x : x z and 2 x 2 are empty?

Solution : Set A consists of those irrational numbers which satisfy 2x 1 0 . If we solve2x 1 0 we get x 1. Clearly 1are not irrational numbers. Therefore A is an empty set.

But B = 2, 1, 0, 1, 2 . B is not an empty set as it has five elements.

Example 15.7 Which of the following sets are singleton ?

A = x : x Z and x 2 0 B = 2y : y R and y 2 0 .

Solution : Set A contains those integers which are the solution of x 2 0 or x = 2. A 2 .A is a singleton set.

B is a set of those real numbers which are solutions of 2y 2 0 or y 2

B 2, 2 Thus, B is not a singleton set.

MATHEMATICS 7

Notes



CHECK YOUR PROGRESS 15.11. Which of the following collections are sets ?

(i) The collection of days in a week starting with S.(ii) The collection of natural numbers upto fifty.(iii) The collection of poems written by Tulsidas.(iv) The collection of fat students of your school.

(2) Insert the appropriate symbol in blank spaces.

If A = 1 ,2 ,3 .

(i) 1...........A (ii) 4........A.3. Write each of the following sets in the Roster form :

(i) A = x : x z and 5 x 0 .

(ii) B = 2x : x R and x 1 0 .

(iii) C = {x : x is a letter of the word banana}.(iv) D = {x : x is a prime number and exact divisor of 60}.

4. Write each of the following sets are in the set builder form ?(i) A = {2, 4, 6, 8, 10} (ii) B = {3, 6, 9,...... }

(iii) C = {2, 3, 5, 7} (iv) D = 2, 2

Are A and B disjoints sets ?5. Which of the following sets are finite and which are infinite ?

(i) Set of lines which are parallel to a given line.(ii) Set of animals on the earth.(iii) Set of Natural numbers less than or equal to fifty.(iv) Set of points on a circle.

6. Which of the following are null set or singleton ?

(i) A = 2x : x R and x is a solution of x 2 0 .

(ii) B = x : x Z and x is a solution of x 3 0 .

(iii) C = 2x : x Z and x is a solution of x 2 0 .

(iv) D = {x : x is a student of your school studying in both the classes XI and XII }7. In the following check whether A = B or A B .

(i) A = {a}, B = {x : x is an even prime number}.(ii) A = {1, 2, 3, 4}, B = {x : x is a letter of the word guava}.

(iii) A = {x : x is a solution of 2x 5x 6 0 },B = {2 , 3}.

MATHEMATICS

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15.4 SUB- SETLet set A be a set containing all students of your school and B be a set containing all students ofclass XII of the school. In this example each element of set B is also an element of set A. Such aset B is said to be subset of the set A. It is written as B A

Consider D ={1, 2, 3, 4,........}E = {..... 3 2, 1, 0, 1, 2, 3, .......}

Clearly each element of set D is an element of set E also D E

If A and B are any two sets such that each element of the set A is an elementof the set B also, then A is said to be a subset of B.

Remarks

(i) Each set is a subset of itself i.e.A A .

(ii) Null set has no element so the condition of becoming a subset is automatically satisfied.Therefore null set is a subset of every set.

(iii) If A Band B A then A = B.

(iv) If A Band A B then A is said to be a proper subset of B and B is said to be asuper set of A. i.e. A B or B A .

Example 15.8 If A = {x : x is a prime number less than 5} and B = {y : y is an even prime number} then is B a proper subset of A ?

Solution : It is given that

A = {2, 3 }, B = {2}.

Clearly B A and B A

We write B A

and say that B is a proper subset of A.

Example 15.9 If A = {1, 2, 3, 4}, B = {2, 3, 4, 5}.

is A B orB A ?

Solution : Here 1 A but1 B A B.

Also 5 B but 5 A B A .

Hence neither A is a subset of B nor B is a subset of A.

Example 15.10 If A = {a, e, i, o, u} B = {e, i, o, u, a }

Is A B orB A or both ?

MATHEMATICS 9

Notes



Solution : Here in the given sets each element of set A is an element of set B also

A B .......... (i)

and each element of set B is an element of set A also. B A ......(ii)

From (i) and (ii)

A = B

15.5 POWER SETLet A = {a, b}

Subset of A are , {a}, {b} and {a, b}.

If we consider these subsets as elements of a new set B (say) then

B = ,{a},{b},{a,b}

B is said to be the power set of A.

Notation : Power set of a set A is denoted by P(A).Power set of a set A is the set of all subsets of the given set.

Example 15.11 Write the power set of each of the following sets :

(i) A = 2x : x R and x 7 0 .

(ii) B = y : y N and1 y 3 .

Solution :

(i) Clearly A = (Null set)

is the only subset of given set

P (A)={ }

(ii) The set B can be written as {1, 2, 3}

Subsets of B are , {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}.

P (B) = , {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3} .

15.6 UNIVERSAL SETConsider the following sets.

A = {x : x is a student of your school}B = {y : y is a male student of your school}C = {z : z is a female student of your school}D = {a : a is a student of class XII in your school}

Clearly the set B, C, D are all subsets of A.

MATHEMATICS

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A can be considered as the universal set for this particular example. Universal set is generallydenoted by U.In a particular problem a set U is said to be a universal set if all the sets in that problem aresubsets of U.

Remarks(i) Universal set does not mean a set containing all objects of the universe.(ii) A set which is a universal set for one problem may not be a universal set for another

problem.

Example 15.12 Which of the following set can be considered as a universal set ?X = {x : x is a real number}Y = {y : y is a negative integer}Z = {z : z is a natural number}

Solution : As it is clear that both sets Y and Z are subset of X. X is the universal set for this problem.

15.7 VENN DIAGRAMBritish mathematician John Venn (1834 1883 AD) introduced the concept of diagrams torepresent sets. According to him universal set is represented by the interior of a rectangle andother sets are represented by interior of circles.For example if U= {1, 2, 3, 4, 5}, A = {2, 4} and B = {1,3}, then these sets can berepresented as

Fig. 15.1

Diagramatical representation of sets is known as a Venn diagram.

15.8 DIFFERENCE OF SETSConsider the sets

A = {1, 2, 3, 4, 5} and B= {2, 4, 6}.A new set having those elements which are in A but not B is said to be the difference of sets Aand B and it is denoted by A B.

A B= {1, 3, 5}Similiarly a set of those elements which are in B but not in A is said to be the difference of B andA and it is devoted by B A.

B A = {6}

MATHEMATICS 11

Notes



In general, if A and B are two sets then

A B = { x : x A and x B}

B A = {x : x B and x A }

Difference of two sets can be represented using Venn diagram as :

or

Fig. 15.2 Fig. 15.3

15.9. COMPLEMENT OF A SETLet X denote the universal set and Y, Z its sub set where

X = {x : x is any member of the family} Y = {x : x is a male member of the family} Z = {x : x is a female member of the family}X Y is a set having female members of the family..X Z is a set having male members of the family..X Y is said to be the complement of Y and is usally denoted by Y' or cY .X Z is said to be complement of Z and denoted by Z' or cZ .

If U is the universal set and A is its subset then the complement of A is a set of those elementswhich are in U which are not in A. It is denoted by A' or cA .

A' = U A = {x : x U and x A}The complement of a set can be represented using Venn diagram as :

Fig. 15.4

Remarks(i) Difference of two sets can be found even if none is a subset of the other but complement

of a set can be found only when the set is a subset of some universal set.

(ii) c U . (iii) cU .

MATHEMATICS

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Example 15.13 Given thatA = {x : x is a even natural number less than or equal to 10}

and B = {x : x is an odd natural number less than or equal to 10}Find (i) A B (ii) B C (iii) is A B=B A ? ?Solution : It is given that

A = {2, 4, 6, 8, 10}, B = {1, 3, 5, 7, 9}Therefore,(i) A B ={2, 4, 6, 8, 10}(ii) B A ={1, 3, 5, 7, 9}(iii) Clearly from (i) and (ii) A B B A.

Example 15.14 Let U be the universal set and A its subset where

U={x : x N and x 10 }

A = {y : y is a prime number less than 10}Find (i) Ac (ii) Represent Ac in Venn diagram.Solution : It is given

U= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.A = { 2, 3, 5, 7}

(i) cA U A = {1, 4, 6, 8, 9, 10}

(ii)

Fig. 15.5

CHECK YOUR PROGRESS 15.21. Insert the appropriate symbol in the blank spaces, given that

A={1, 3, 5, 7, 9}

(i) .............A (ii) {2, 3, 9}..........A

(iii) 3............A (iv) 10...................A2. Given that A = {a, b}, how many elements P(A) has ?

3. Let A = { , {1} ,{2}, {1,2}}

Which of the following is true or false ?

MATHEMATICS 13

Notes



(i) {1,2} AA (ii) A .

4. Which of the following statements are true or false ?(i) Set of all boys, is contained in the set of all students of your school.(ii) Set of all boy students of your school, is contained in the set of all students of your

school.(iii) Set of all rectangles, is contained in the set of all quadrilaterals.(iv) Set of all circles having centre at origin is contained in the set of all ellipses having

centre at origin.5. If A = { 1, 2, 3, 4, 5 } , B= {5, 6, 7} find (i) A B (ii) B A.6. Let N be the universal set and A, B, C, D be its subsets given by

A = {x : x is a even natural number}B = {x : x N and x is a multiple of 3}C = {x : x N and x 5}D = {x : x N and x 10}Find complements of A, B, C and D respectively.

15.10. INTERSECTION OF SETSConsider the sets

A = {1, 2, 3, 4} and B = { 2, 4, 6}It is clear, that there are some elements which are common to both the sets A and B. Set of thesecommon elements is said to be interesection of A and B and is denoted by A B .Here A B = {2, 4 }If A and B are two sets then the set of those elements which belong to both the sets is said to bethe intersection of A and B. It is devoted by A B .

A B = {x : x A and x B}

A B can be represented using Venn diagram as :

Fig. 15.6

Remarks

If A B then A and B are said to be disjoint sets. In Venn diagram disjoint sets canbe represented as

Fig.15.7

MATHEMATICS

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Example 15.15 Given thatA = {x : x is a king out of 52 playing cards}

and B = { y : y is a spade out of 52 playing cards}Find (i) A B (ii) Represent A B by using Venn diagram .Solution : (i) As there are only four kings out of 52 playing cards, therefore the set A has onlyfour elements. The set B has 13 elements as there are 13 spade cards but out of these 13 spadecards there is one king also. Therefore there is one common element in A and B. A B ={ King of spade}.(ii)

Fig. 15.8

15.11 UNION OF SETS

Consider the following examples :(i) A is a set having all players of Indian men cricket team and B is a set having all players of

Indian women cricket team. Clearly A and B are disjoint sets. Union of these two sets isa set having all players of both teams and it is denoted by A B .

(ii) D is a set having all players of cricket team and E is the set having all players of Hocketyteam, of your school. Suppose three players are common to both the teams then unionof D and E is a set of all players of both the teams but three common players to bewritten once only.

If A and B are only two sets then union of A and B is the set of those elements which belong toA or B.

In set builder form :

A B = {x : x A or x B}OR

A B = {x : x A B or x B A or x A B }

A B can be represented using Venn diagram as :

Fig. 15.9 Fig. 15.10

MATHEMATICS 15

Notes



n(A B) n(A B) n(B A) n(A B) .

or n(A B) n(A) n(B) n(A B)

where n A B stands for number of elements in A B so on.

Example 15.16 A = x : x Z and 5

B = {y : y is a prime number less than 10}Find (1) A B (ii) represent A B using Venn diagram.Solution : We have,

A={1, 2, 3, 4, 5} B = {2, 3, 5, 7}.

A B = {1, 2, 3, 4, 5, 7}.

(ii)

Fig.15.11

CHECK YOUR PROGRESS 15.31. Which of the following pairs of sets are disjoint and which are not ?

(i) {x : x is an even natural number}, {y : y is an odd natural number}

(ii) {x : x is a prime number and divisor of 12}, { y : y N and 3 y 5 }

(iii) {x :x is a king of 52 playing cards}, { y : y is a diamond of 52 playing cards}(iv) {1, 2, 3, 4, 5}, {a, e, i, o, u}

2. Find the intersection of A and B in each of the following :

(i) A = {x : x Z}, B= {x : x N } (ii) A = {Ram, Rahim, Govind, Gautam}

B = {Sita, Meera, Fatima, Manprit}

3. Given that A = {1, 2, 3, 4, 5}, B={5, 6, 7, 8, 9, 10}

find (i) A B (ii) A B .

4. If A ={x : x N }, B={y : y z and 10 y 0 } find A B and write youranswer in the Roster form as well as set-builder form.

5. If A ={2, 4, 6, 8, 10}, B {8, 10, 12, 14}, C ={14, 16, 18, 20}.

Find (i) A B C (ii) A B C .6. Let U = {1, 2, 3,.......10}, A = {2, 4, 6, 8, 10}, B = { 1, 3, 5, 7, 9, 10}

Find (i) A B ' (ii) A B ' (iii) (B')' (iv) (B A )'.

7. Draw Venn diagram for each the following :

MATHEMATICS

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(i) A B when B A (ii) A B when A and B are disjoint sets.(iii) A B when A and B are neither subsets of each other nor disjoint sets.

8. Draw Venn diagram for each the following :(i) A B when A B. (ii) A B when A and B are disjoint sets.(iii) A B when A and B are neither subsets of each other nor disjoint sets.

9. Draw Venn diagram for each the following :(i) A B and B A when A B.

(ii) A B and B A when A and B are disjoint sets.

(iii) A B and B A when A and B are neither subsets of each other nor disjoint sets.

15.12 CARTESIAN PRODUCT OF TWO SETSConsider two sets A and B where

A={1, 2}, B= {3, 4, 5}.Set of all ordered pairs of elements of A and Bis {(1,3), (1,4), (1,5), (2,3), (2,4), (2,5)}This set is denoted by A × B and is called the cartesian product of sets A and B.i.e. A×B ={(1, 3), (1, 4),(1, 5),(2, 3),(2, 4),(2, 5)}Cartesian product of B sets and A is denoted by B×A.In the present example, it is given by

B×A = {(3, 1),(3, 2),(4, 1),(4, 2),(5, 1),(5, 2)}Clearly A×B B×A.In the set builder form :

A×B = {(a,b) : a A and b B}

B×A = {(b,a) : b B and a A }

Note : If A or B or A,B

then A B B A .

Example 15.17

(1) Let A={a,b,c}, B={d,e}, C={a,d}.

Find (i) A×B(ii) B×A (iii) A×(B C ) (iv) (A C) B

(v) (A B) C (vi)A (B C) .

Solution : (i) A×B ={(a, d),(a, e), (b, d), (b, e), (c, d), (c, e)}.(ii) B×A = {(d, a),(d, b), (d, c), (e, a) (e, b),(e, c)}.(iii) A = {a, b, c}, B C ={a,d,e}.

MATHEMATICS 17

Notes



A × (B C ) ={(a, a),(a, d),(a, e),(b, a),(b, d),(b, e), (c, a),(c, d),(c, e).

(iv) A C = {a}, B={d, e}.

(A C )×B={(a, d), (a, e)}

(v) A B = , c={a,d}, A B c

(vi) A = {a,b,c}, B C {e}.

A (B C) {(a,e),(b,e),(c,e)} .

15.13 RELATIONSConsider the following example :

A={Mohan, Sohan, David, Karim}B={Rita, Marry, Fatima}

Suppose Rita has two brothers Mohan and Sohan, Marry has one brother David, and Fatimahas one brother Karim. If we define a relation R " is a brother of" between the elements of A andB then clearly.Mohan R Rita, Sohan R Rita, David R Marry, Karim R Fatima.After omiting R between two names these can be written in the form of ordered pairs as :(Mohan, Rita), (Sohan, Rita), (David, Marry), (Karima, Fatima).The above information can also be written in the form of a set R of ordered pairs as

R= {(Mohan, Rita), (Sohan, Rita), (David, Marry), Karim, Fatima}

Clearly R A B, i.e.R {(a,b):a A,b B and aRb}

If A and B are two sets then a relation R from A toB is a sub set of A×B.

If (i) R , R is called a void relation.

(ii) R=A×B, R is called a universal relation.(iii) If R is a relation defined from A to A, it is called a relation defined on A.

(iv) R = (a,a) a A , is called the identity relation.

15.13.1 Domain and Range of a RelationIf R is a relation between two sets then the set of its first elements (components) of all theordered pairs of R is called Domain and set of 2nd elements of all the ordered pairs of R iscalled range, of the given relation.Consider previous example given above.

Domain = {Mohan, Sohan, David, Karim}Range = {Rita, Marry, Fatima}

Example 15.18 Given that A = {2, 4, 5, 6, 7}, B = {2, 3}.

R is a relation from A to B defined by

MATHEMATICS

Notes


18


R = {(a, b) : a A, b B and a is divisible by b}

find (i) R in the roster form (ii) Domain of R (iii) Range of R(iv) Repersent R diagramatically.

Solution : (i) R = {(2, 2), (4, 2), (6, 2), (6, 3)}(ii) Domain of R = {2, 4, 6}(iii) Range of R = {2, 3}

(iv)

Fig. 15.12

Example 15.19 If R is a relation 'is greater than' from A to B, whereA= {1, 2, 3, 4, 5} and B = {1,2,6}.

Find (i) R in the roster form. (ii) Domain of R (iii) Range of R.Solution :(i) R = {(3, 1), (3, 2), (4, 1), (4, 2), (5, 1), (5, 2)}(ii) Domain of R = {3, 4, 5}(iii) Range of R = {1, 2}

CHECK YOUR PROGRESS 15.41. Given that A = {4, 5, 6, 7}, B = {8, 9}, C = {10}

Verify that A (B C) = (A B) (A C) .

2. If U is a universal set and A, B are its subsets.Where U= {1, 2, 3, 4, 5}.

A = {1,3,5}, B = {x : x is a prime number} find A' × B'3. If A = {4, 6, 8, 10}, B = {2, 3, 4, 5}

R is a relation defined from A to B where

R= {(a, b) : a A, b B and a is a multiple of b}

find (i)R in the Roster form (ii) Domain of R (iii) Range of R .4. If R be a relation from N to N defined by

R= {(x,y) : 4x y 12, x , y N }

find (i) R in the Roster form (ii) Domain of R (iii) Range of R.5. If R be a relation on N defined by

R={ 2(x,x ) : x is a prime number less than 15}

MATHEMATICS 19

Notes



Find (i) R in the Roster form (ii) Domain of R (iii) Range of R

6. If R be a relation on set of real numbers defined by R={(x,y) : 2 2x y 0} Find

(i) R in the Roster form (ii) Domain of R (iii) Range of R.

15.14 DEFINITION OF A FUNCTIONConsider the relation

f : {(a, 1), (b, 2), (c, 3), (d, 5)}In this relation we see that each element of A has a uniqueimage in BThis relation f from set A to B where every element of Ahas a unique image in B is defined as a function from A toB. So we observe that in a function no two orderedpairs have the same first element.

We also see that an element B, i.e., 4 which does not have its preimage in A. Thus here:

(i) the set B will be termed as co-domain and(ii) the set {1, 2, 3, 5} is called the range.From the above we can conclude that range is a subset of co-domain.Symbolically, this function can be written as

f : A B or A f B

Example 15.20 Which of the following relations are functions from A to B. Write their

domain and range. If it is not a function give reason ?

(a) (1, 2),(3,7),(4, 6),(8,1) , A 1,3,4,8 , B 2,7, 6,1,2

(b) (1,0),(1 1),(2,3),(4,10) , A 1,2,4 , B 0, 1,3,10

(c) (a,b),(b,c),(c,b),(d,c) , A a,b,c ,d ,e B b,c

(d) (2,4),(3,9),(4,16),(5,25),(6,36 , A 2,3,4,5,6 , B 4,9,16,25,36

(e) (1, 1),(2, 2),(3, 3),(4, 4),(5, 5) ,A 0,1,2,3,4,5 ,

B 1, 2, 3, 4, 5

(f)1 3 1

sin , , cos , , tan , , cot , 36 2 6 2 6 63 ,

A sin ,cos ,tan ,cot6 6 6 6

1 3 1B , , , 3 ,1

2 2 3

(g) (a,b),(a,2),(b,3),(b,4) , A a ,b , B b,2,3,4 .

Fig.15.13

MATHEMATICS

Notes


20


Solution :(a) It is a function.

Domain= 1,3,4,8 , Range 2,7, 6,1

(b) It is not a function. Because Ist two ordered pairs have same first elements.(c) It is not a function.

Domain= a,b,c ,d A , Range = b ,c

(d) It is a function.

Domain 2,3,4,5,6 , Range 4,9,16,25,36

(e) It is not a function .

Domain 1,2,3,4,5 A , Range 1, 2, 3, 4, 5

(f) It is a function .

Domain sin ,cos ,tan ,cot6 6 6 6 , Range

1 3 1, , , 3

2 2 3

(g) It is not a function.First two ordered pairs have same first component and last two ordered pairs have also samefirst component.

Example 15.21 State whether each of the following relations represent a function or not.

(a) (b)

Fig.15.14 Fig.15.15

(c) (d)

Fig. 15.16 Fig.15.17

MATHEMATICS 21

Notes



Solution :

(a) f is not a function because the element b of A does not have an image in B.

(b) f is not a function because the element c of A does not have a unique image in B.

(c) f is a function because every element of A has a unique image in B.

(d) f is a function because every element in A has a unique image in B.

Example 15.22 Which of the following relations from R . R are functions?

(a) y 3x 2 (b) y x 3 (c) 2y 2x 1

Solution :

(a) y 3x 2

Here corresponding to every element x R, a unique element y R.

It is a function.

(b) y x 3 .

For any real value of x we get more than one real value of y.It is not a function.

(c) 2y 2x 1

For any real value of x, we will get a unique real value of y.It is a function.

CHECK YOUR PROGRESS 15.51. Which of the following relations are functions from A to B ?

(a) (1, 2),(3,7),(4, 6),(8,11) , A 1,3,4,8 , B 2,7, 6,11

(b) (1,0),(1, 1),(2,3),(4,10) , A 1,2,4 , B 1,0, 1,3,10

(c) (a,2),(b,3),(c,2),(d,3) , A a ,b ,c ,d ,B 2,3

(d) (1,1),(1,2),(2,3),( 3,4) , A 1,2, 3 , B 1,2 ,3 ,4

(e)1 1 1

2, , 3, ,...., 10, ,2 3 10

A 1,2 ,3 ,4 ,1 1 1

B , ,...,2 3 11

(f) 1,1 , 1,1 , 2,4 , 2,4 , A 0,1, 1,2, 2 , B 1, 4

2. Which of the following relations represent a function ?

MATHEMATICS

Notes


22


(a) (b)

Fig. 15.18 Fig. 15.19

(c) (d)

Fig. 15.20 Fig. 15.21

3. Which of the following relations defined from R R are functions ?(a) y 2x 1 (b) y x 3 (c) y 3x 1 2(d) y x 1

4. Write domain and range for each of the following functions :

(a) 2 ,2 , 5, 1 , 3 ,5

(b)1 1 1

3, , 2, , 1,2 2 2

(c) 1,1 , 0,0 , 2,2 , 1, 1

(d) Deepak,16 , Sandeep,28 , Rajan,24

5. Write domain and range for each of the following mappings :(a) (b)

Fig. 15.22 Fig. 15.23

MATHEMATICS 23

Notes



(c) (d)

Fig. 15.24 Fig. 15.25

(e)

Fig. 15.26

15.14.1 Some More Examples on Domain and RangeLet us consider some functions which are only defined for a certain subset of the set of realnumbers.

Example 15.23 Find the domain of each of the following functions :

(a)1yx

(b) 1y

x 2(c) 1y

(x 2)(x 3)

Solution : The function 1yx

can be described by the following set of ordered pairs.

1 1......., 2, , 1, 1 , 1,1 2, ,....

2 2Here we can see that x can take all real values except 0 because the corresponding image, i.e.,

10

is not defined.

Domain R 0 [Set of all real numbers except 0]

Note : Here range = R { 0 }

(b) x can take all real values except 2 because the corresponding image, i.e., 12 2

does

not exist.Domain R 2

(c) Value of y does not exist for x 2 and x 3

Domain R 2,3

MATHEMATICS

Notes


24


Example 15.24 Find domain of each of the following functions :

(a) y x 2 (b) y 2 x 4 x

Solution :(a) Consider the function y x 2

In order to have real values of y, we must have x 2 0

i.e. x 2Domain of the function will be all real numbers 2 .

(b) y 2 x 4 x

In order to have real values of y, we must have 2 x 4 x 0

We can achieve this in the following two cases :

Case I : 2 x 0 and 4 x 0

x 2 and x 4

Domain consists of all real values of x such that 4 x 2

Case II : 2 x 0 and 4 x 0

2 x and x 4 .But, x cannot take any real value which is greater than or equal to 2 and less than or equal to

4 .From both the cases, we have

Domain = 4 x 2 x R

Example 15.25 For the function

f x y 2x 1, find the range when domain 3, 2, 1,0,1,2,3 .

Solution : For the given values of x, we have

f 3 2 3 1 5

f 2 2 2 1 3

f 1 2 1 1 1

f 0 2 0 1 1

f 1 2 1 1 3

f 2 2 2 1 5

f 3 2 3 1 7

The given function can also be written as a set of ordered pairs.

MATHEMATICS 25

Notes



i.e., 3, 5 , 2, 3 , 1, 1 , 0,1 1, 3 , 2,5 3,7

Range 5, 3, 1,1,3,5,7

Example 15.26 If f x x 3, 0 x 4,find its range.

Solution : Here 0 x 4or 0 3 x 3 4 3

or 3 f x 7

Range = f x : 3 f x 7

Example 15. 27 If 2f x x , 3 x 3 ,find its range.

Solution : Given 3 x 3

or 20 x 9 or 0 f x 9

Range = f x : 0 f x 9

CHECK YOUR PROGRESS 15.61. Find the domain of each of the following functions x R :

(a) (i) y = 2x (ii) y = 9x + 3 (iii) 2y x 5

(b) (i)1y

3x 1(ii) 1y

4x 1 x 5

(iii)1y

x 3 x 5(iv) 1y

3 x x 5

(c) (i) y 6 x (ii) y 7 x

(iii) y 3x 5

(d) (i) y 3 x x 5 (ii) y x 3 x 5

(iii)1y

3 x 7 x(iv)

1yx 3 7 x

2. Find the range of the function, given its domain in each of the following cases.

(a) (i) f x 3x 10 , x 1, 5, 7, 1, 2

(ii) 2f x 2x 1, x 3, 2, 4, 0

(iii) 2f x x x 2 , x 1, 2, 3, 4, 5

MATHEMATICS

Notes


26


(b) (i) f x x 2 , 0 x 4 (ii) f x 3x 4 , 1 x 2

(c) (i) 2f x x , 5 x 5 (ii) f x 2x , 3 x 3

(iii) 2f x x 1 , 2 x 2 (iv) f x x , 0 x 25

(d) (i) f x x 5 , x R (ii) f x 2x 3 , x R

(iii) 3f x x , x R (iv) 1f xx

, x : x 0

(v) 1f xx 2

, x : x 1 (vi) 1f x3x 2

, x : x 0

(vii) 2f xx

, x : x 0 (viii)xf x

x 5, x : x 5

15.15 CLASSIFICATION OF FUNCTIONSLet f be a function from A to B. If every element of the set B is the image of at least one elementof the set A i.e. if there is no unpaired element in the set B then we say that the function f mapsthe set A onto the set B. Otherwise we say that the function maps the set A into the set B.Functions for which each element of the set A is mapped to a different element of the set B aresaid to be one-to-one.

One-to-one function

Fig.15.27

The domain is A,B,C

The co-domain is 1,2 ,3 ,4

The range is 1,2,3A function can map more than one element of the set A to the same element of the set B. Sucha type of function is said to be many-to-one.

Many-to-one function

Fig. 15.28

MATHEMATICS 27

Notes



The domain is A,B,C

The co-domain is 1,2 ,3 ,4

The range is 1, 4A function which is both one-to-one and onto is said to be a bijective function.

Fig. 15.29 Fig. 15.30

Fig. 15.31 Fig. 15.32

Fig. 15.29 shows a one-to-one function mapping A,B,C into 1,2 ,3 ,4 .

Fig. 15.30 shows a one-to-one function mapping A,B,C onto 1,2,3 .

Fig. 15.31shows a many-to-one function mapping A,B,C into 1,2 ,3 ,4 .

Fig. 15.32 shows a many-to-one function mapping A,B,C onto 1, 2 .

Function shown in Fig. 15.30 is also a bijective Function.Note : Relations which are one-to-many can occur, but they are not functions. The followingfigure illustrates this fact.

Fig. 15.33

Example 15.28 Without using graph prove that the function

F : R R defiend by f x 4 3x is one-to-one.

Solution : For a function to be one-one function

1 2 1 2 1 2F x f x x x x , x domain

MATHEMATICS

Notes


28


Now 1 2f x f x gives

1 2 1 24 3x 4 3x or x xF is a one-one function.

Example 15.29 Prove that

F : R R defined by 3f x 4x 5 is a bijection

Solution : Now 1 2 1 2f x f x x , x Domain

3 31 24x 5 4x 5

3 31 2x x

3 31 2x x 0

2 22 1 1 1 2 2x x x x x x 0

1 2x x or2 2

1 1 2 2x x x x 0 (rejected). It has no real value of 1x and 2x .

F is a one-one function.

Again let y x where y codomain, x domain.

We have 3y 4x 5 or1 / 3y 5x

4

For each y codomain x domain such that f x y .

Thus F is onto function. F is a bijection.

Example 15.30 Prove that F : R R defined by 2F x x 3 is neither one-one noronto function.

Solution : We have 1 2 1 2F x F x x , x domain giving

2 2 2 21 2 1 2x 3 x 3 x x

or 2 21 2 1 2 1 2x x 0 x x or x x

or F is not one-one function.

Again let y F x where y codomain

x domain.

2y x 3 x y 3

y 3 no real value of x in the domain.

F is not an onto finction.

MATHEMATICS 29

Notes



15.16 GRAPHICAL REPRESENTATION OF FUNCTIONSSince any function can be represented by ordered pairs, therefore, a graphical representation ofthe function is always possible. For example, consider 2y x .

2y x

x 0 1 1 2 2 3 3 4 4y 0 1 1 4 4 9 9 16 16

Fig. 15.34Does this represent a function?Yes, this represent a function because corresponding to each value of x a unique value of y..

Now consider the equation 2 2x y 252 2x y 25

x 0 0 3 3 4 4 5 5 3 3 4 4y 5 5 4 4 3 3 0 0 4 4 3 3

Fig. 15.35

MATHEMATICS

Notes


30


This graph represents a circle.Does it represent a function ?No, this does not represent a function because corresponding to the same value of x, theredoes not exist a unique value of y.

CHECK YOUR PROGRESS 15.71. (i) Does the graph represent a function?

Fig. 15.36(ii) Does the graph represent a function ?

Fig. 15.372. Which of the following functions are into function ?

(a)

Fig.15.38

(b) f : N N, defined as 2f x x

Here N represents the set of natural numbers.

(c) f : N N, defined as f x x

3. Which of the following functions are onto function if f : R R

MATHEMATICS 31

Notes



(a) f x 115x 49 (b) f x x

4. Which of the following functions are one-to-one functions ?

(a) f : 20,21,22 40,42,44 defined as f x 2x

(b) f : 7,8,9 10 defined as f x 10

(c) f : I R defined as 3f x x

(d) f : R R defined as 4f x 2 x

(d) f : N N defined as 2f x x 2x

5. Which of the following functions are many-to-one functions ?

(a) f : 2, 1,1,2 2,5 defined as 2f x x 1

(b) f : 0,1,2 1 defined as f x 1

(c)

Fig.15.39

(d) f : N N defined as f x 5x 7

6. Draw the graph of each of the following functions :

(a) 2y 3x (b) 2y x (c) 2y x 2

(d) 2y 5 x (e) 2y 2x 1 (f) 2y 1 2x

7. Which of the following graphs represents a function ?

(a) (b)

Fig. 15.40 Fig. 15.41

MATHEMATICS

Notes


32


(c) (d)

Fig. 15.42 Fig. 15.43

(e)

Fig .15.44

Hint : If any line || to y-axis cuts the graph in more than one point, graph does not represent afunction.

15.17 SOME SPECIAL FUNCTIONS15.17.1 Monotonic FunctionLet F : A B be a function then F is said to be monotonic on an interval (a,b) if it is eitherincreasing or decreasing on that interval.For function to be increasing on an interval (a,b)

1 2 1 2 1 2x x F x F x x x a, b

and for function to be decreasing on (a,b)

1 2 1 2 1 2x x F x F x x x a, b

A function may not be monotonic on the whole domain but it can be on different intervals of thedomain.

Consider the function F : R R defined by 2f x x .

Now 1 2x , x 0,

1 2 1 2x x F x F x

F is a Monotonic Function on 0, .

MATHEMATICS 33

Notes



( ∵ It is only increasing function on this interval)

But 1 2x , x ,0

1 2 1 2x x F x F x

F is a Monotonic Function on , 0

( ∵ It is only a decreasing function on this interval)Therefore if we talk of the whole domain given function is not monotonic on R but it is monotonicon , 0 and 0, .

Again consider the function F : R R defined by 3f x x .

Clearly 1 2x x domain

1 2 1 2x x F x F x

Given function is monotonic on R i.e. on the whole domain.

15.17.2 Even FunctionA function is said to be an even function if for each x of domain

F( x) F(x)

For example, each of the following is an even function.

(i) If 2F x x then 2 2F x x x F x

(ii) If F x cosx then F x cos x cosx F x

(iii) If F x x then F x x x F x

Fig. 15.45

The graph of this even function (modulus function) is shown in the figure above.

Observation Graph is symmetrical about y-axis.

MATHEMATICS

Notes


34


15.17.3 Odd FunctionA function is said to be an odd function if for each x

f x f x

For example,

(i) If 3f x x

then 3 3f x x x f x

(ii) If f x sin x

then f x sin x sin x f x

Graph of the odd function y = x is given in Fig.15.46

Observation

Graph is symmetrical about origin.

15.17.4 Greatest Integer Function (Step Function)

f x x which is the greatest integer less than or equal to x.

f x is called Greatest Integer Function or Step Function. Its graph is in the form of steps, asshown in Fig. 16.47.

Let us draw the graph of y x , x R

x 1, 1 x 2

x 2, 2 x 3

x 3, 3 x 4

x 0, 0 x 1

x 1, 1 x 0

x 2, 2 x 1� Domain of the step function is the set of real numbers.� Range of the step function is the set of integers.

15.17.5 Polynomial FunctionAny function defined in the form of a polynomial is called a polynomial function.For example,

(i) 2f x 3x 4x 2

(ii) 3 2f x x 5x x 5

Fig. 15.46

Fig. 15.47

MATHEMATICS 35

Notes



(iii) f x 3

are all polynomial functions.

Note : Functions of the type f x k , where k is a constant is also called a constant function.

15.17.6 Rational Function

Function of the type g xf x

h x, where h x 0 and g x and h x are polynomial

functions are called rational functions.

For example,2x 4f x , x 1x 1

is a rational function.

15.17.7 Reciprocal Function

Functions of the type 1y , x 0x

is called a reciprocal function.

15.17.8 Exponential FunctionsA swiss mathematician Leonhard Euler introduced a number e in the form of an infinite series. Infact

1 1 1 1e 1 ..... .....1 2 3 n .....(1)

It is well known that the sum of its infinite series tends to a finite limit (i.e., this series is convergent)and hence it is a positive real number denoted by e. This number e is a transcendental irrationalnumber and its value lies between 2 an 3.Consider now the infinite series

2 3 nx x x x1 ..... .....1 2 3 n

It can be shown that the sum of its infinite series also tends to a finite limit, which we denote byxe . Thus,

2 3 nx x x x xe 1 ..... .....

1 2 3 n ......(2)

This is called the Exponential Theorem and the infinite series is called the exponential series.We easily see that we would get (1) by putting x = 1 in (2).

The function xf x e , where x is any real number is called an Exponential Function.

The graph of the exponential functionxy e

MATHEMATICS

Notes


36


is obtained by considering the following important facts :(i) As x increases, the y values are increasing very rapidly, whereas as x decreases, the y

values are getting closer and closer to zero.

(ii) There is no x-intercept, since xe 0 for any value of x.

(iii) The y intercept is 1, since 0e 1 and e 0 .

(iv) The specific points given in the table will serve as guidelines to sketch the graph of xe(Fig. 15.48).

x 3 2 1 0 1 2 3xy e 0.04 0.13 0.36 1.00 2.71 7.38 20.08

Fig. 15.48

If we take the base different from e, say a, wewould get exponential function

xf x a ,provbided a 0, a 1.For example, we may take a = 2 or a = 3 and getthe graphs of the functions

xy 2 (See Fig. 15.49)

and xy 3 (See Fig. 15.50)

Fig. 15.49

MATHEMATICS 37

Notes



Fig. 15.50 Fig. 15.51

15.17.9 Logarithmic FunctionsConsider now the function

xy e .....(3)We write it equivalently as

ex log y

Thus, ey log x .....(4)

is the inverse function of xy eThe base of the logarithm is not written if it is eand so elog x is usually written as log x.

As xy e and y log x are inverse functions,their graphs are also symmetric w.r.t. the liney xThe graph of the function y log x can be

obtained from that of xy e by reflecting it inthe line y = x.

Note(i) The learner may recall the laws of indices which you have already studied in the Secondary

Mathematics :If a > 0, and m and n are any rational numbers, then

m n m na a a

Fig. 15.52

MATHEMATICS

Notes


38


m n m na a a

nm mna a

0a 1(ii) The corresponding laws of logarithms are

a a alog mn log m log n

a a amlog log m log nnna alog m n log m

ab

a

log mlog mlog b

or b a blog m log mlog a

Here a, b > 0, a 1, b 1.

CHECK YOUR PROGRESS 15.81. Tick mark the correct statement.

(i) Function 4 2f x 2x 7x 9x is an even function.

(ii) Odd function is symmetrical about y-axis.

(iii) 1 / 2 3 5f x x x x is a polynomial function.

(iv) x 3f x3 x

is a rational function for all x R.

(v)5f x

3is a constant function.

(vi) 1f xx

Domain of the function is the set of real numbers except 0.(vii) Greatest integer function is neither even nor odd.

2. Which of the following functions are even or odd functions ?

(a)2x 1f x

x 1(b)

2

2xf x

5 x(c)

21f x

x 5

(d)3

2f xx

(e)2xf x

x 1 (f) 5f x

x 5

MATHEMATICS 39

Notes



(g) x 3f x3 x

(h) 3f x x x

3. Draw the graph of the function y x 2 .

4. Specify the following functions as polynomial function, rational function, reciprocal functionor constant function.

(a) 8 7 5y 3x 5x 8x (b)2

3x 2xy

x 2x 3, 3x 2x 3 0

(c) 23y

x, x 0 (d)

2x 1y 3 , x 0x

(e)1y 1 , x 0x

(f)2x 5x 6y , x 2

x 2

(g)1y9

.

15.18 COMPOSITION OF FUNCTIONSConsider the two functions given below:

y 2x 1, x 1,2,3

z y 1, y 3,5,7Then z is the composition of two functions x and y because z is defined in terms of y and y interms of x.Graphically one can represent this as given below :

Fig. 15.53

The composition, say, gof of function g and f is defined as function g of function f.

If f : A B and g : B C

then g o f : A to C

Let f x 3x 1 and 2g x x 2

Then fog x f g x

MATHEMATICS

Notes


40


2f x 2

23 x 2 1 23x 7 (i)

and gof x g f x

g 3x 1

23x 1 2 29x 6x 3 (ii)

Check from (i) and (ii), iffog = gof

Evidently, fog gof

Similarly, fof x f f x f 3x 1 [Read as function of function f ].

3 3x 1 1

9x 3 1 9x 4

2gog x g g x g x 2 [ Read as function of function g ]

22x 2 2

4 2x 4x 4 24 2x 4x 6

Example 15.31 If f x x 1 and 2g x x 2 , calculate fog and gof.

Solution : fog x f g x

2f x 2

2x 2 1

2x 3

gof x g f x

g x 1

2 2x 1

x 1 2 = x + 3.

Here again, we see that fog gof

Example 15.32 If 3f x x , f : R R

1g x , g : R 0 R 0x

MATHEMATICS 41

Notes



Find fog and gof.Solution : fog x f g x

1f

x

3

31 1x x

gof x g f x

3g x 31x

Here we see that fog = gofNote : We observe from Example 15.31 and Example 15.32 that fog and gof may or maynot be equal.

CHECK YOUR PROGRESS 15.9

1. Find fog, gof, fof and gog for the following functions :

2f x x 2,1g x 1 , x 1.

1 x2. For each of the following functions write fog, gof, fof and gog.

(a) 2f x x 4 , g x 2x 5

(b) 2f x x , g x 3

(c) f x 3x 7 ,2g x , x 0x

3. Let f x | x |, g x x . Verify that fog gof.

4. Let 2f x x 3, g x x 2

Prove that fog gof and 3 3f f g f

2 2

5. If 2f x x , g x x . Show that fog gof.

6. Let13

1f x | x |, g x x , h x ; x 0.x

Find (a) fog (b) goh (c) foh (d) hog (e) fogoh

1fogoh x f g h x f g

xHint :

MATHEMATICS

Notes


42


15.19 INVERSE OF A FUNCTION(A) Consider the relation

Fig. 15.54

This is a many-to-one function. Now let us find the inverse of this relation.Pictorially, it can be represented as

Fig 15.55

Clearly this relation does not represent a function. (Why ?)(B) Now take another relation

Fig.15.56

It represents one-to-one onto function. Now let us find the inverse of this relation, which isrepresented pictorially as

Fig. 15.57

MATHEMATICS 43

Notes



This represents a function.(C) Consider the relation

Fig. 15.58

Ir represents many-to-one function. Now find the inverse of the relation.Pictorially it is represented as

Fig. 15.59

This does not represent a function, because element 6 of set B is not associated with anyelement of A. Also note that the elements of B does not have a unique image.(D) Let us take the following relation

Fig. 15.60

It represent one-to-one into function.Find the inverse of the relation.

Fig. 15.61

MATHEMATICS

Notes


44


It does not represent a function because the element 7 of B is not associated with any elementof A. From the above relations we see that we may or may not get a relation as a function whenwe find the inverse of a relation (function).We see that the inverse of a function exists only if the function is one-to-one onto functioni.e. only if it is a bijective function.

CHECK YOUR PROGRESS 15.101 (i) Show that the inverse of the function

y 4x 7 exists.

(ii) Let f be a one-to-one and onto function with domain A and range B. Write thedomain and range of its inverse function.

2. Find the inverse of each of the following functions (if it exists) :

(a) f x x 3 x R

(b) f x 1 3x x R

(c) 2f x x x R

(d)x 1f x , x 0 x R

x

LET US S

� Set is a well defined collection of objects.� To represent a set in Roster form all elements are to be written but in set builder form a

set is represented by the common property.� If the elements of a set can be counted then it is called a finite set and if the elements

cannot be counted, it is infinite.� If each element of set A is an element of set B also then A is called sub set of B.

� For two sets A and B, A B is a set of those elements which are in A but not in B.� Complement of a set A is a set of those elements which are in the universal set but not in

A. i.e. cA U A� Intersection of two sets is a set of those elements when belong to both the sets.� Union of two sets is a set of those elements which belong to either of the two sets.� Cartesian product of two sets A and B is the set of all ordered pairs of the elements of A

and B. It is denoted by A B. i.e.A B a, b : a A and b B .

� Relation is a sub set of A×B where A and B are sets.i.e. R A B a ,b : a A and b B and aRb

LET US SUM UP

MATHEMATICS 45

MODULE -IVFunctions

Notes


● Function is a special type of relation.● Functions f : A → B is a rule of correspondence from A to B such that to every element

of A ∃ a unique element in B.● Functions can be described as a set of ordered pairs.● Let f be a function from A to B.

Domain : Set of all first elements of ordered pairs belonging to f.Range : Set of all second elements of ordered pairs belonging to f.

● Functions can be written in the form of equations such as y = f(x)where x is independent variable, y is dependent variable.Domain : Set of independent variable.Range : Set of dependent variable.Every equation does not represent a function.

● Vertical line test : To check whether a graph is a function or not, we draw a lineparallel to y-axis. If this line cuts the graph in more than one point, we say that graphdoes not represent a function.

● Let f be a function from a set A to a set B. Symbolically we can write it asf : A → B(i) If every element of B is not an image of some element of A then f is said to be intofunction. In this case, range ⊂ co-domain.(ii) If range = co-domain, then f is said to be onto function.(iii) If distinct elements of set A have distinct images in set B then f is called one-to-onefunction.(iv) If many elements in the domain of a function have the same image element in therange, then the function is called many-to-one function.

● Horizontal Line Test : To check the one-to-oneness of a function, draw a line parallel,to x-axis. If it cuts the graph at one point, we say that it is one-to-one function.

● A function is said to be monotonic on an interval if it is either increasing or decreasing onthat interval.

● A function is called even function if f(x) = f(−x), and odd function iff(−x) = −f(x), x −x ∈ Df

● Inverse of a function exists if it is one-to-one and onto.

SUPPORTIVE WESITES

● http://www.wikipedia.org

● http://mathworld.walfram.com

1. Which of the following statements are true or false :

46 MATHEMATICS

MODULE -IVFunctions

Notes


(i) {1, 2, 3} = { 1, {2}, 3} (ii) {1, 2, 3 } = {3, 1, 2}

(iii) {a, e, 0 } = {a, b, c} (iv) { φ} = { }

2. Write the set in Roster form represented by the shaded portion in the following.

(i) A = {1, 2, 3, 4, 5 }

B = { 5, 6, 7, 8, −9}

Fig. 15.62

(ii) A = {1, 2, 3, 4, 5, 6 }

B = { 2, 6, 8, 10, 12 }

Fig. 15.63

3. Represent the follwoing using Venn diagram.

(i) (A ∪ B)' provided A and B are not disjoint sets.

(ii) (A ∩ B)' provided A and B are disjoint sets.

(iii) (A − B)' provided A and B are not disjoint sets.

4. Let U = {1,2,3,4,5,6, 7,8,9,10}, A{2,4,6,8}, B = {1,3,5, 7}

Verify that

(i) (A ∪ B)' = A' ∩ B'

(ii) ( A ∩ B )' = A' ∪ B'

(iii) (A − B) ∪ (B - A) = ( A ∪ B) − (A ∩ B) .

5. Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10},

A = {1, 3, 5, 7, 9} . B = {2, 4, 6, 8, 10} ,

C = {1, 2, 3}.

Find (i) A' ∩ (B − C). (ii) A ∪ ( B ∪ C)

(iii) A' ∩ (B ∪ C)' (iv) ( A ∩ B)' ∪ C'

6. What does the shaded ponion represent in each of the following Venn diagrams:

(i) (ii)

Fig. 15.64 Fig. 15.65

MATHEMATICS 47

MODULE -IVFunctions

Notes


(iii) (iv)

Fig. 15.66 Fig. 15.67

7. Draw Venn diagram for the following:

(i) A' ∩ (B ∪ C ) (ii) A' ∩ (C − B)

Where A, B, C are not disjoint sets and are subjects of the universal set U.

8. Given A = {a, b, c,}, B = { 2, 3 } .

Find the number of relations from A to B.

9. Given that A = { 7,8,9} , B = {9, 10,11 }, C = {11, 12} verify that

(i) A × (B ∩ C) = (A × B) ∩ (A × C)

(ii) A × (B ∪ C) = (A × B) ∪ (A × C)

10. Which of the following equations represent functions?

In each of the case x ∈ R

(a) y = 2x 3 4

,x4 5x 5

+ ≠−

(b) y = 3

,x 0x

≠

(c) y = 2

3,x 4, 4

x 16≠ −

−(d) y = x 1, x 1− ≥

(e) y = 2

1

x 1+(f) x2 + y2 = 25

11. Write domain and range of the following functions :

ft : {(0, 1),(2, 3),(4, 5)(6, 7), ...... (100, 101)}

f2 : {( −2, 4), (−4, 16), (−6, 36)}

f3 : 1 1 1

(1,1), , 1 , , 1 , 1 , .......2 3 4

− −

f4 : {...... (3, 0), (−1, 2) , ( 4, −1) }

f5 : {...... (−3, 3),(−2, 2),(−1, 1) (0, 0) (1, l), (2, 2), ...... }

12. WritedomainofthefollowingfiJnctions:

48 MATHEMATICS

MODULE -IVFunctions

Notes


(a) f(x) = x3 (b) f(x) = 2

1

x 1−

(c) f(x) = 3x 1+ (d) f (x) = 1

(x 1) (x 6)+ +

(e) f(x) = 1

(x 1) (5x 5)− −

13. Write range of each of the following fimctions :

(a) y = 3x + 2, x ∈ R (b) y = 1

x 2−, x ∈ R − {2}

(c) y = x 1

x 1

−+ , x ∈ {0, 2, 3, 5, 7, 9}

(d) y = 2

x, x ∈ R+(All non-negative real values)

14. Draw the graph of each of the following fimctions :

(a) y = x2 + 3, x ∈ R (b) y = 1

x 2−, x ∈ R − {2}

(c) y = x 1

x 1

−+ , x ∈ {0, 2, 3, 5, 7,9 } (d) y =

2

x, x ∈ R+.

15. Which of the following graphs represent a fimction ?

(a) (b)

Fig. l5.68 Fig.l5.69

(c) (d)

Fig. 15.70 Fig. 15.71

MATHEMATICS 49

MODULE -IVFunctions

Notes


(e) (f)

Fig. 15.72 Fig. 15.73

(g) (h)

Fig. 15.74 Fig. 15.75

16. Which of the following functions are one-to-one or many-to-one function?

(a) y = 3x + 5, x ∈ R

(b) y = 1 − 2x2, x ∈ R

(c) f : {(1, 2), (2, 2),(3, 2), (4, 2)}

(d) f : {1, 2, 3} → { 4, 5, 6, 8} such that f(1) = 4, f(2) = 6, f(3) = 8.

17. Which of the following functions are rational functions ?

(a) f(x) = 2x 3

x 2

−+

, x ∈ R − { −2} (b) f (x) = x

x, x ∈ R4

(c) f(x) = 2

x 2

x 4x 4

++ +

, x ∈ R − { −2} (d) y = x, x ∈ R

1 8. Which of the following functions are polynomial functions ?

(a) f(x) = x2 + 3 x + 2 (b) f(x) = (x + 2)2

(c) f(x) = 3 − x + 2x3 − x4 (d) f(x) = x + x − 5, x > 0

(e) f(x) = 2x 4− , x ∉ (− 2, 2)

19. Which of the following functions are even or odd functions ?

(a) f(x) = 29 x− x ∈ [ − 3, 3] (b) f(x) = 2

2

x 1

x 1

−+

(c) f(x) = |x | (d) f ( x) = x − x5

50 MATHEMATICS

MODULE -IVFunctions

Notes


(e) (f )

Fig. 15.76 Fig. 15.77

(g)

Fig. 15.78

20. Write for each of the following functions fog, gof, fof, gogo

(a) f(x) = x3 g (x) = 4x − 1

(b) f(x) = 2

1

x, x ≠ 0 g (x) = x2− 2x + 3

(c) f(x) = x 4− , x > 4 g(x) = x − 4

(d) f(x) = x2 − 1 g (x) = x2 + 1

21. (a) Let f(x) = |x |, g(x) = 1

x, x ≠ 0, h(x) = x1/3. Find fogoh

(b) f(x) = x2 + 3, g(x) = 2x2 + 1. Find fog (3) and gof(3).

22. Which of the following equations describe a function whose inverse exists :

(a) f(x) = |x | (b) f(x) = x , x > 0

(c) f (x) = x2 − 1, x > 0 (d) f(x) = 3x 5

4

−

(e) f (x) = 3x 1

x 1

+− x ≠ 1

MATHEMATICS 51

MODULE -IVFunctions

Notes


ANSWERS


1. (i), (ii), (iii) are sets.

2. (i) ∈ (ii) ∉

3. (i) A = { − 5, − 4, − 3, − 2, − 1, 0}

(ii) B = { − 1, 1}, (iii) C = { a, b, n} (iv) D = {2, 3, 5} .

4. (i) A = {x : x is a even natural number less than or equal to ten}.

(ii) B = { x : x ∈ N and x is a multible of 3}.

(iii) C = {x : x is a prime number less than 10}.

(iv) D = { x : x ∈ R and x is a solution of x2 − 2 = 0}.

5. (i) Infinite (ii) Finite (iii) Finite

(iv) Infinite

6. (i) Null (ii) Singleton (iii) Null (iv) Null

7. (i) A ≈ B (ii) A ≈ B (iii) A = B .


1. (i) ⊂ (ii) ⊄ (iii) ∈ (iv) ∉

2. 4

3. (i) False (ii)True

4. (i) False (ii) True (iii) True (iv) False

5. (i) {1, 2, 3, 4} (ii) {6, 7} .

6. A e = {x : x is a odd natural number}

Be = { x : x ∈ N and x is not a multiple of 3 }

CC = {1, 2, 3, 4} .

De = {11, 12, 13, ...... }

52 MATHEMATICS

MODULE -IVFunctions

Notes



1. (i) Disjoint (ii) Not disjoint (iii) Not disjoint (iv) Disjoint

2. (i) {x : x ∈ N} (ii) 4>

3. (i) {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}(ii) {5}

4. Roster from {-10, -9, -8,0, 1, 2, 3, }

Set builder from {x : x ∈ Z and -10 < x < oo}

5. (i) { x : x is a even natural number less than equal to 20}. (ii) 4>

6. (i) 4> (ii) { 1, 2, 3, 4, 5, 6, 7, 8, 9 }

(iii) { 1, 3, 5, 7,9,10 } (iv) { 2, 4, 6, 8, 10 }

7. (i) (ii)

Fig. 15.79 Fig. 15.80

(iii)

Fig. 15.81

8. (i) (ii)

Fig.l5.82 Fig. l5.83

(iii)

Fig. 15.84

MATHEMATICS 53

MODULE -IVFunctions

Notes


9. (i)

A − B = φ B - A' = Shaded Portion

Fig. 15.85

(ii)

A − B = A B − A = B

Fig. 15.86

(iii)

A − B = Shaded Portion B − A = Shaded Portion

Fig. 15.87

2. {( 2, 1 ) , ( 4, 1 ) , (2,4) , ( 4, 4)} .

3. (i) R = { ( 4, 2) ,.( 4, 4) , ( 6, 2), ( 6, 3), (8,2), (8,4), (10,2), (10,5) } .

(ii) Domain of R = { 4, 6, 8, 10 } (iii) Rangeof R = {2, 3, 4, 5}.

4. (i) R = {(1, 8), (2, 4)} (ii) Domain of R = {1, 2}.

(iii) Range of R = {l, 2 }{ 8, 4}

5. (i) R = {(2,4),(3,9),(5, 25), (7,49), (11, 121), (13, 169)}

Domain of R = {2, 3, 5, 7, 11, 13}, Range of R = { 4, 9, 25, 49, 121, 169}.

6. (i) Domain of R = φ (ii) Domain of R = φ (iii) Range of R = φ

54 MATHEMATICS

MODULE -IVFunctions

Notes



1. (a), (c), (f) 2. (a), (b) 3. (a), (d)

4. (a) Domain = { }2, 5, 3 , Range = {2, -1, 5}

(b) Domain = {-3, -2, -1 },Range = 1

2

(c) Domain = {1, 0, 2, − 1} , Range = {1, 0, 2, − 1}

(d) Domain = {Deepak, Sandeep, Rajan} , Range = {16, 28, 24} .

5. (a) Domain = {1, 2, 3}, Range = {4, 5, 6}

(b) Domain= {1, 2, 3}, Range = {4}

(c) Domain = {1, 2, 3 } , Range = {1, 2, 3}

(d) Domain = {Gagan, Ram, Salil}, Range = {8, 9, 5}

(e) Domain = {a.b, c, d} , Range = {2, 4}


1. (a) (i) Domain = Set of real numbers. (ii) Domain = Set of real numbers.

(iii) Domain = Set of a real numbers.

(b) (i) Domain = R 1

3 −

(ii) Domain = R 1

, 54

− −

(iii) Domain = R − {3, 5 } (iv) Domain = R − {3, 5}

(c) (i) Domain= {x ∈ R: x < 6} (ii) Domain = {x ∈ R : x > −7}

(iii) Domain = 5

x : x R,x3

∈ ≥ −

(d) (i) Domain = { x : x ∈ R and 3 < x < 5}

(ii) Domain = { x : x ∈ R x > 3, x < −5}

(iii) Domain = {x: x ∈ R x > −3, x < −7}

(iv) Domain = { x : x ∈ R x > 3, x < −7}

MATHEMATICS 55

MODULE -IVFunctions

Notes


2. (a) (i) Range = {13, 25, 31, 7, 4} (ii) Range = {19, 9, 33, 1}

(iii) Range = { 2, 4, 8, 14, 22 }

(b) (i) Range = {f(x) : −2< f(x) < 2} (ii) Range = {f(x) : 1 < f(x) < 10}

(c) (i) Range = {f(x) : 1 < f(x) < 25} (ii) Range = {f(x): −6< f(x) < 6}

(iii) Range = {f(x) : 1 < f(x) < 5} (iv) Range = {f( x) : 0 < f(x) < 5}

(d) (i) Range = R (ii) Range = R (iii) Range = R

(iv) Range = {f(x) : f(x) < 0}

(v) Range = { f ( x ) : −1 < f(x) < 0}

(vi) Range= {f(x): 0.5 < f(x) < 0} (vii) Range = {f(x) : f(x) > 0}

(viii) Range: All values of f(x) except values at x = −5.


l. (i) No. (ii) Yes 2. (a), (b)

3. (a) 4. (a), (c), (e) 5. (a), (b)

6. (a) (b)

Fig. 15.88 Fig. 15.89

(c) (d)

Fig. 15.90 Fig. 15.91

56 MATHEMATICS

MODULE -IVFunctions

Notes


(e) (f )

Fig. 15.92 Fig. 15.93

7. (c), (d) and (e).


1. v, vi, vii are true statements.

(i), (ii), (iii), (iv) are incorrect statement.

2. (b) (c) are even functions.

(d) (e) (h) are odd functions.

3.

Fig. 15.94

4. (a) Polynomial function (b) Rational ftmction.

(c) Rational function. (d) Rational function.

(e) Rational function. (f ) Rational function.

MATHEMATICS 57

MODULE -IVFunctions

Notes


(g) Constant function

5. No


1. (i) fog = 2

2

x2

(x 1)+

−(ii) gof =

2

2

x 2

x 1

++

(iii) fof = x4 + 4x2 + 6 (iv) gog = x

2. (a) fog = 4x2 + 20x + 21

gof = 2x2 − 3

fof = x4 − 8x2 + 12

gog = 4x + 15

(b) fog = 9, gof = 3, fof = x4, gog = 3

(c) fog = 6 7x

x

−, gof =

2

3x 7−, fof = 9x − 28, gog = x

6. (a) fog = 13x (b) goh =

13x (c) foh =

1

x

(d) hog = 6 7x

x

−(e) fogoh(l) = 1


1. (ii) Domain is B. Range is A.

2. (a) f −l(x) = x − 3 (b) f −l(x) = 1 x

3

−

(c) Inverse does not exist. (d) f−l(x) = 1

x 1−

TERMINAL EXERCISE

1. (i) False (ii) True (iii)False (iv) False

2. (i).{ 1, 2, 3, 4, 6, 7, 8, 9} (ii) {8, 10, 12 }

58 MATHEMATICS

MODULE -IVFunctions

Notes


(i)

Fig. 15.95

(ii)

Fig. 15.96

(iii)

Fig. 15.97

5. (i) { 4, 6, 8, 10 } (ii) {1, 2, 3, 4, 5,6, 7, 8, 9, 10}.

(iii) φ (iv) {1,2,3,4,5,6, 7,8,9,10}.

6. (i) ( A ∩ B) ∩ ( B ∩ C) (ii) (A ∩ B) ∩ C

(iii) [(A ∪ B) ∪ C]' (iv) A u ( B ∩ C) .

7. (i)

Fig. 15.98

(ii)

Fig. 15.99

MATHEMATICS 59

MODULE -IVFunctions

Notes


8. 26 i.e., 64.

10. (a), (b), (c), (d), (e) are functions.

11. f1 − Domain = {0, 2, 4, 6, ....., 100}.

Range = {1, 3, 5, 7, ..... 101}.

f2 − Domain = {− 2, − 4, − 6, ........ } .

Range = {4, 16, 36, .........}.

f3 − Domain = 1 1 1

1, , , , ......2 3 4

−

Range = {l, − l} .

f4 − Domain = {3, − 1, 4}.

Range = {0, 2, -l}.

f5 − Domain = {......., 3, − 2, − 1, 0, 1, 2, 3, .........}.

Range = {0, 1, 2, 3, .........} .

12. (a) Domain = R . (b) Domain = R − { − 1, 1}.

(c) Domam = 1

x x R3

≥ − ∀ ∈

(d) Domain = x > − 1, x < − 6. (e) Domam = 5

x , x 12

≥ ≤

13. (a) Range = R (b) Range = All values of y except at x = 2.

(c) Range = 1 1 2 3 4

1, , , , ,3 2 3 4 5

−

(d) Range = All values of y for x > 0

15. (a), (c), (e), (t), (h). Use hint given in check your progress 15.7, Q. NO.7 for the

solution.

16. One-to-one functions: (a), (d) Many-to-one functions: (b), (c)

17. (a), (c) 18. (a), (b), (c)

19. Even functions: (a), (b), (c), (t), (g)

Odd functions : (d), (e)

60 MATHEMATICS

MODULE -IVFunctions

Notes


20. (a) fog = ( 4x − 1)3, gof = 4x3 − 1, fof = x9, gog = 16x − 5

(b) fog = 2 2

1

(x 2x 3)− + , gof = 4 2

4

3x 2x 1

x

− +

fog = x4, gog = x4 − 4x3 + 4x2

(c) fog = x 8− gof = x 4 4− −

fof = x 4 4− − , gog = x - 8

(d) fog = x4 + 2x2, gof = x4− 2x2 + 2,

fof = x4 − 2x2, gog = x4 + 2x2 + 2

21. (a) 13

1

x

(b) (fog) (3) = 364, (gof)(3) = 289

22. (c), (d), (e)

60/1

Functions

15.11 Real Variable Function

1. A function f : A→B is called a real variable function ifA R⊆ .

2. A function f : A→B is called real valued function if B R⊆ .

3. A function f : A→B is called real function ifA R⊆ , B R⊆ .

Example: ax (a > 0), sin x, log x these are all real functions.

15.11.1 nth root of a non-negative real number

Let xbe a non-negative real number and n be a positive integer then there exists a unique non-

negative real number y such that nsuch that yn = x. This number y is called the nth root of x and is

denoted as 1nx (or) n x

Example 1: The domain of the real valued function 2 2( ) = −f x a x (a > 0) is [ , ]−a a .

[since 2 2 Ra x− ∈ (a > 0)

2 2 2 20⇔ − ≥ ⇔ ≤a x x a

| |⇔ ≤ ⇔ − ≤ ≤x a a x a]

15.11.2 Algebra of real valued functions

If f and g are real valued functions with domains A and B respectively, then both f and g are

defined on A B∩ whenA B∩ = φ.

i) (f + g) (x) = f(x) + g(x) defined on A B∩

ii) ( f g) (x) = f(x) g(x) defined on A B∩

60/2

iii) ( )( )

( )

=

f f xx

g g x∀ x ∈ E, E = { A B / ( ) 0}x g x∈ ∩ ≠

iv) Let f : A →R and n ∈ N, we define | f | and f n

on A by | | ( ) | ( ) | ( ) ( ( )) A= = ∀ ∈n nf x f x f x f x x .

v) If E = { A / ( ) 0}x f x∈ ≥ ≠ φ , then we define f on E by ( ) ( )=f x f x , for all x ∈ E.

Find the domains of the following real valued functions.

Example 2 : 2

1( )

6 5=

− −f x

x x

Solution: 2

1 1( ) R

( 1)(5 )6 5= = ∈

− −− −f x

x xx x

⇔ (x − 1) (x − 5) ≠ 0

⇔ x ≠ 1, 5

∴ Domain of f is R − {1, 5}.

Example 3: ( ) ( 2)( 3)= + −f x x x

Solution : ( 2)( 3) R+ − ∈x x

⇔ (x + 2) (x − 3) > 0

⇔ x < −2 or x > 3

⇔ ( , 2] [3, )∈ −∞ − ∪ ∞x

= R − (−2, 3).

Example 4: 2

2

1( ) 1

3 2= − +

− +f x x

x x

Sol : f (x) = 2

2

11 R

3 2− + ∈

− +x

x x

⇔ x2 − 1 > 0, x2 − 3x + 2 > 0

60/3

⇔ (x + 1) (x − 1) > 0 and (x − 1)(x − 2) > 0

⇔ x ∈ (−∞ , −1) ∪ [1, ∞ ] and x ∈ (−∞ , 1) ∪ (2, ∞ )

⇔ x ∈ R − (−1, 1) ∩ R − (1, 2)

⇔ x ∈ R − {(−1, 1) ∪ (1, 2)}, ⇔ x + R − [−1, 2] = (−∞ , −1)] ∪ (2, ∞ )

Domain of f is ´ (−∞ , 1) ∪ (2, ∞ ) = R − (−1, 2).

Exmaple 5(i) : f = {(4, 5) (5, 6), (6, −4)} and g = {(4, −4), (6, 5), (8, 5)} find (i) ( f − g)

Sol : f − g = {(4, 5 + 4), (6, −4 −5)}

= {(4, 9), (6, −9)}.

Example 5(ii) : Find 2f + 4g

Sol : Domain of 2f ´ {4, 5, 6} Domain of 4g ´ {4, 6, 8}

But 2f = {(4, 10) (5, 12) (6, −8)}

4g = {(4, −16) (6, 20) (8, 20)}

Domain of 2f + 4g ́ {4, 6}

2f + 4g = {(4, 10 − 16), (6, −8 + 20)}

= {(4, −6), (6, 12)}.

Example 5(iii) : Find fg

Sol : Domain of fg A B∩ ´ {4, 6}

fg ´ {(4, (5) (−4)), (6, (−4)(−5)}

= {(4, −20), (6, 20)}.

Example 5(iv) : Find f

g

Sol : Domain of f

g ´ {4, 6}

60/4

5 44, , 6,

4 5

− − ∴ = f

g

Example 5(v) : Find f 2

Sol : Domain of f 2 ´ A = {4, 5, 6}

f2 = {(4, 25), (5, 36), (6, 16)}.

Find the Domains and ranges of the following real valued functions.

Example 6: 2

( )2

xf x

x

+=−

Sol : 2( ) R 2 0 2

2

xf x x x

x

+= ∈ ⇔ − ≠ ⇔ ≠−

R {2}x⇔ ∈ −

Domain of f is R − {2}

Let f (x) = y

2 2( 1)

2 1

x yy x

x y

+ −⇒ = ⇒ =− +

x is not defined for y + 1 = 0 i.e., when y = −1

∴ range of f ´ R − { −1}.

Example 7: 2( ) 9f x x= −

Sol : 2 29 R 9 0x x− ∈ ⇔ − ≥

(3 ) (3 ) 0x x⇔ + − ≥

[ 3, 3]x⇔ ∈ −

∴ Domain of f = [−3, 3]

60/5

y = f (x) 29 Rx y⇒ = − ∈

29 0 (3 )(3 ) 0y y y⇔ − ≥ ⇔ + − ≥

∴ − 3< y < 3 but f (x) attains only non negative values.

∴ range of f = [0, 3].

15.12 Types of Functions

15.12.1 Implicit function : y is said to be an implicit function of x if it is given in the form of

f(x, y) = 0.

Ex: x2 + xy + y2 − 2 = 0.

15.12.2 Polynomial function : If n is a non negative integer, a1,a2, ..., an are real number (at least

one 0ia ≠ ) then the function f defined on R by f (x) = a0 + a1x + a2x2 + ....+ anx

n for all Rx∈

is called a polynomial function.

15.12.3 (i) Rational function :

If φ(x), ψ(x) are polynomial functions and ψ(x) ≠ 0 for all Rx∈ then the function( )

( )

x

x

φψ

is

called a rational function.

The domain of this function ́ R − {x / ψ(x) = 0}

Ex: 2

2

5 6( )

4 3

xf x

x x

−=+ +

15.12.3 (ii) Even and odd functions

Let A be a non empty subset of R such that− Ax∈ for all Ax∈ and t f : A → R

(i) If f (−x) = f(x) for every x in A then f is called an even function.

(ii) If f(−x) = −f(x) for every x in A then f is called an odd function.

60/6

Example:

(i) f(x) = x2 + 5sin2x, g(x) = cos x are even functions

(ii) f (x) = sin x, g(x) = x3 + 5x are odd functions.

15.12.4 Algebraic functions

Binary operations like addition, subtraction, multiplication, division and extraction of square root

etc. are called algebric operations. A function obtained by applying a finite number of algebraic opera-

tions on polynomial function is called an algebraic function.

Ex: 2( ) 2 3f x x x= + is an algebraic function.

15.12.5 Exponential function

The function ax when1 0a≠ > and x is rational, is called exponential function.

The domain of this functuon (ax) is R i.e., ( , )−∞ ∞ .

The range is R+ ´ (0, )∞

Ex: If : R Rf +→ can be defined as f (x) = 2x is anexponential function.

Logarithmic function

15.12.6 If 1,a ≠ a > 0 given x > 0 then the function f(x) = logax is called lagarithmic function. The

domain of f is (0, )∞ and its rangle is R i.e. ( , )−∞ ∞ .

Ex: If +: R Rf → then the function f(x) = log2x is called lagarithmic function.

15.12.7 Greatest integer function

The function :R Rf → defined byf (x) = [x] ∀ Rx ∈ , for any real number x, we denote by

[x] the greatest integer less than or equal to x, is called the greatest integer function.

The domain of this function is R and the range is the setz of all integers.

60/7

0 < x < 1, ∀ x values f (x) = 0

1 < x < 2, ∀ x values f(x) = 1

−1 < x < 0, ∀ x values f(x) = −1

−2 < x < 1, ∀ x values f(x) = −2

-----------------------

...............................................................

Example 8 : If f(x) = x2, g(x) = |x| then find the fillowing functions :

1) f + g 2) f − g 3) fg 4) f 2

Example 8.1: Sol: f(x) = x2, g(x) = |x| = 2

2

, 0

, 0

x x x

x x x

+ ≥

− <, domain f ´ domaing ´ R

Hence the domain of all functions (i) through iv is R.

22

2

, 0( )( ) ( ) ( ) | |

, 0

x x xf g x f x g x x x

x x x

+ ≥+ = + = + =− <

.

Example 8.2: Sol: 2

2

2

, 0( )( ) ( ) ( ) | |

, 0

x x xf g x f x g x x x

x x x

− ≥− = − = − =+ <

Example 8.3: Sol: 3

2

3

, 0( )( ) ( ) ( ) | |

, 0

x xfg x f x g x x x

x x

≥= = = − <

Example 8.4 Sol: 2 2 2 2 4( ) ( ( )) ( )f x f x x x= = =

Example 9 : Determine whether the following function is even or odd.

Sol: 2( ) log ( 1)f x x x= + +

2( ) log( 1)f x x x⇒ − = − + +

60/8

2 2

2

( 1)( 1)( ) log

( 1)

x x x xf x

x x

+ + − + + − = + +

( )2 2 12

2

1log log 1

1

x xx x

x x

− + −= = + + + +

( )2( ) log 1 ( )f x x x f x− = − + + = −

∴ f is an odd function

Example 10: Find the domain of the following real valued function 10

1( ) 2

log (1 )f x x

x= + +

−

Sol:: 10

1( ) 2 R

log (1 )f x x

x= + + ∈

−

2 0, 1 0, 1 1x x x⇔ + ≥ − > − ≠

2, 1 , 0x x x⇔ ≥ − > ≠

[ 2, ) ( , 1) {0}x⇔ ∈ − ∞ ∩ −∞ −

⇔ Domain of f is [−2, 1) − {0}.

Check your progress - 1

1. Find the domain of real valued function 22 5 7

( )( 1)( 2)( 3)

x xf x

x x x

− +=− − −

.

2. Find the domain of 2( ) 4f x x x= − .

3. Find the domain of 2( ) 25f x x= − .

4. Find the range of the following functions.

i)2

sin [ ]

1 [ ]

x

x

π+

ii) 2 4

2

x

x

−−

60/9

5. If f(x) = 2x − 1, g(x) = x2 then find the following.

i) ( )f

xg

ii) ( f + g + 2) (x)

6. If f = {(1, 2) (2, −3), (3, −1)} then find

i) 2f ii) f 2 iii) f

Answers

1. R − {1, 2, 3}

2. [0, 4]

3. R − (−5, 5)

4. i) {0} ii) R − {4}

5. i) 2

2 1x

x

−ii) (x + 1)2

6. i) {(1, 4), (2, −6), (3, −2)}

ii) {(1, 4) (2, 9), (3, 1)}

iii) { }(1, 2)

Let us sumup

1. A function : A Bf → is a real valued function if A R,⊆ B R⊆ .

2. If yn = x then y is nth root of x. This can be written as 1nx or n x

3. If f, g are real functions with domains A, B two sets then bothf, g are defined on A B∩ .

4. ( ) ( ) ( )f g x f x g x± = ± domain =A B∩

5. ( ) ( ) ( )f g x f x g x= domain ́ A B∩

60/10

6. ( )

( )

f f xx

g g x

=

– Domain ´ { x/x ∈ A B∩ , g(x) ≠ 0}

7. Let A be a non empty subset of R such that −x ∈ A ∀ x ∈ A and : A Rf → .

(i) If f (−x) = f (x) ∀ x ∈ A then f is called even.

(ii) If f(−x) = −f (x) ∀ x ∈ A then f is called odd.

8. If f (a) = ax (a > 0) then f (x) is Exponential function.

9. If f (x) = logax (a ≠ 1, x > 0) thenf (x) is logarithmic function.

Web sites

� http://www.wikipedia.org

� http://mathworld.wolfram.com

MATHEMATICS 61

Notes


Trigonometric Functions-I

16

TRIGONOMETRIC FUNCTIONS - I

We have read about trigonometric ratios in our earlier classes. Recall that we defined theratios of the sides of a right triangle as follows :

c a csin , cos , tanb b a

and cosec b b a, sec , cotc a c

We also developed relationships between thesetrigonometric ratios as

2 2sin cos 1, 2 2sec 1 tan , 2 2cosec 1 cotWe shall try to describe this knowledge gained so far in terms of functions, and try todevelop this lesson using functional approach.In this lesson, we shall develop the science of trigonometry using functional approach.We shall develop the concept of trigonometric functions using a unit circle. We shalldiscuss the radian measure of an angle and also define trigonometric functions of the typey = sin x, y = cos x, y = tan x, y = cot x, y = sec x, y = cosec x, y = a sin x, y = b cos x,etc., where x, y are real numbers.

We shall draw the graphs of functions of the type

y = sin x, y = cos x, y = tan x, y = cotx, y = secx, and y = cosecx y = a sin x,y = a cos x.

OBJECTIVESAfter studying this lesson, you will be able to :

define positive and negative angles;define degree and radian as a measure of an angle;convert measure of an angle from degrees to radians and vice-versa;

Fig.16.1

MATHEMATICS

Notes


62


state the formula r� where r and have their usual meanings;

solve problems using the relation r� ;

define trigonometric functions of a real number;draw the graphs of trigonometric functions; andinterpret the graphs of trigonometric functions.

EXPECTED BACKGROUND KNOWLEDGEDefinition of an angle.Concepts of a straight angle, right angle and complete angle.Circle and its allied concepts.

Special products : 2 2 2a b a b 2ab , 3 3 3a b a b 3ab a b

Knowledge of Pythagoras Theorem and Py thagorean numbers.

16.1 CIRCULAR MEASURE OF ANGLEAn angle is a union of two rays with the common end point. An angle is formed by the rotationof a ray as well. Negative and positive angles are formed according as the rotation is clock-wise or anticlock-wise.

16.1.1 A Unit CircleIt can be seen easily that when a line segment makes one complete rotation, its end pointdescribes a circle. In case the length of the rotating line be one unit then the circle described willbe a circle of unit radius. Such a circle is termed as unit circle.

16.1.2 A RadianA radian is another unit of measurement of an angle other than degree.A radian is the measure of an angle subtended at the centre of a circle by an arc equal in lengthto the radius (r) of the circle. In a unit circle one radian will be the angle subtended at the centreof the circle by an arc of unit length.

Fig. 16.2

Note : A radian is a constant angle; implying that the measure of the angle subtended byan are of a circle, with length equal to the radius is always the same irrespective of theradius of the circle.

MATHEMATICS 63

Notes



16.1.3 Relation between Degree and Radian

An arc of unit length subtends an angle of 1 radian. The circumference 2 ( r 1)∵ subtendan angle of 2 radians.

Hence 2 radians = 360° radians = 180°

2 radians = 90°

4 radians = 45°

1 radian = 3602

=180

or 1° = 2

360 radians =

180 radians

Example 16.1 Convert

(i) 90° into radians (ii) 15° into radians

(iii)6

radians into degrees. (iv)10

radians into degrees.

Solution :

(i) 1° = 2

360 radians

90° = 2 90360

radiansor 90° = 2

radians

(ii) 15° = 2 15360

radians or 15° = 12

radians

(iii) 1 radian = 3602

6 radians =

3602 6

6 radians = 30°

(iv)10

radians = 3602 10

10 radians = 18°

MATHEMATICS

Notes


64


CHECK YOUR PROGRESS 16.11. Convert the following angles (in degrees) into radians :

(i) 60° (ii) 15° (iii) 75° (iv) 105° (v) 270°2. Convert the following angles into degrees:

(i)4

(ii)12

(iii)20

(iv)60

(v)23

3. The angles of a triangle are 45°, 65° and 70°. Express these angles in radians

4. The three angles of a quadrilateral are 6

,3

,23

. Find the fourth angle in radians.

5. Find the angle complementary to 6

.

16.1.4 Relation Between Length of an Arc and Radius of the CircleAn angle of 1 radian is subtended by an arc whose length is equal to the radius of the circle. Anangle of 2 radians will be substened if arc is double the radius.

An angle of 2½ radians willbe subtended if arc is 2½ times the radius.

All this can be read from the following table :

Length of the arc (l) Angle subtended at thecentre of the circle (in radians)

r 1

2r 2

(2½)r 2½

4r 4

Therefore,r�

or r�

where r = radius of the circle,

= angle substended at the centre in radians

and � = length of the arc.

The angle subtended by an arc of a circle at the centre of the circle is given by the ratio of thelength of the arc and the radius of the circle.

Note : In arriving at the above relation, we have used the radian measure of the angle

and not the degree measure. Thus the relationr� is valid only when the angle is

measured in radians.

MATHEMATICS 65

Notes



Example 16.2 Find the angle in radians subtended by an arc of length 10 cm at the centre ofa circle of radius 35 cm.

Solution : � = 10cm and r 35cm.

r� radians or

1035

radians

or27

radians

Example 16.3 If D and C represent the number of degrees and radians in an angle prove that

D C180

Solution : 1 radian = 3602

or180

C radians = 180C

Since D is the degree measure of the same angle, therefore,180D C

which impliesD C

180 Example 16.4 A railroad curve is to be laid out on a circle. What should be the radius of acircular track if the railroad is to turn through an angle of 45° in a distance of 500m?

Solution : Angle is given in degrees. To apply the formula r ,� must be changed toradians.

45 45180

radians ....(1)

=4

radians

� = 500 m ....(2)

� = r gives r �

500r m

4

[using (1) and (2)]

=4500 m

MATHEMATICS

Notes


66


= 2000 ×0.32 m1

0.32

= 640 mExample 16.5 A train is travelling at the rate of 60 km per hour on a circular track. Through

what angle will it turn in 15 seconds if the radius of the track is 56

km.

Solution : The speed of the train is 60 km per hour. In 15 seconds, it will cover

60 15 km60 60

=1 km4

We have, � =1 km4

and r =5 km6

14 radians5r6

�

3 radians10

CHECK YOUR PROGRESS 16.21. Express the following angles in radians :

(a) 30° (b) 60° (c) 150°2. Express the following angles in degrees :

(a)5

(b) 6

(c) 9

3. Find the angle in radians and in degrees subtended by an arc of length 2.5 cm at thecentre of a circle of radius 15 cm.

4. A train is travelling at the rate of 20 km per hour on a circular track. Through what angle

will it turn in 3 seconds if the radius of the track is1

12 of a km?.

5. A railroad curve is to be laid out on a circle. What should be the radius of the circulartrack if the railroad is to turn through an angle of 60° in a distance of 100 m?

6. Complete the following table for l, r, having their usual meanings.

MATHEMATICS 67

Notes



l r

(a) 1.25m ....... 135°

(b) 30 cm .......4

(c) 0.5 cm 2.5 m ........(d) ......... 6 m 120°

(e) ......... 150 cm15

(f) 150 cm 40 m ........

(g) ........ 12 m6

(h) 1.5 m 0.75 m ........(i) 25 m ........ 75°

16.2 TRIGONOMETRIC FUNCTIONSWhile considering, a unit circle you must have noticed that for every real number between 0 and2 , there exists a ordered pair of numbers x and y. This ordered pair (x, y) represents thecoordinates of the point P.

(i) (iii)

(ii) (iv)Fig. 16.3

MATHEMATICS

Notes


68

Trigonometric Functions-IIf we consider 0 on the unit circle, we will have a point whose coordinates are (1,0).

If2

, then the corresponding point on the unit circle will have its coordinates (0,1).

In the above figures you can easily observe that no matter what the position of the point,corresponding to every real number we have a unique set of coordinates (x, y). The values ofx and y will be negative or positive depending on the quadrant in which we are considering thepoint.Considering a point P (on the unit circle) and the corresponding coordinates (x, y), we definetrigonometric functions as :

sin y , cos x

ytan (forx 0)x

,xcot (for y 0)y

1sec (for x 0)x

,1cosec (for y 0)y

Now let the point P move from its original position in anti-clockwise direction. For variouspositions of this point in the four quadrants, various real numbers will be generated. Weesummarise, the above discussion as follows. For values of in the :I quadrant, both x and y are positve.II quadrant, x will be negative and y will be positive.III quadrant, x as well as y will be negative.IV quadrant, x will be positive and y will be negative.or I quadrant II quadrant III quadrant IV quadrant

All positive sin positive tan positive cos positivecosec positive cot positive sec positive

Where what is positive can be rememebred by :All sin tan cos

Quardrant I II III IVIf (x, y) are the coordinates of a point P on aunit circle and , the real number generatedby the position of the point, then sin = y andcos = x. This means the coordinates of thepoint P can also be written as (cos , sin )

From Fig. 16.5, you can easily see that thevalues of x will be between 1 and + 1 as Pmoves on the unit circle. Same will be true fory also.Thus, for all P on the unit circle

P (cos , sin )� �

Fig. 16.4

MATHEMATICS 69

Notes


Trigonometric Functions-I1 x 1 and 1 y 1

Thereby, we conclude that for all real numbers 1 cos 1 and 1 sin 1

In other words, sin and cos can not be numerically greater than 1

Example 16.6 What will be sign of the following ?

(i) sin718

(ii) cos49

(iii) tan59

Solution :

(i) Since 718

lies in the first quadrant, the sign of sin718

will be posilive.

(ii) Since 49

lies in the first quadrant, the sign of cos49

will be positive.

(iii) Since 59

lies in the second quadrant, the sign of tan59

will be negative.

Example 16.7 Write the values of (i) sin 2

(ii) cos 0 (iii) tan2

Solution : (i) From Fig.16.5, we can see that the coordinates of the point A are (0,1)

sin2

=1 , as sin y

Fig.16.5

(ii) Coordinates of the point B are (1, 0)cos 0 1, as cos x

(iii)sin 12tan

2 0cos2

which is not defined

Thus tan2

is not defined.

MATHEMATICS

Notes


70


Example 16.8 Write the minimum and maximum values of cos .

Solution : We know that 1 cos 1

The maximum value of cos is 1 and the minimum value of cos is 1.

CHECK YOUR PROGRESS 16.31. What will be the sign of the following ?

(i) cos3

(ii)5tan6

(iii)2sec3

(iv)35sec18

(v)25tan18

(vi)3cot4

(vii)8cosec3

(viii)7cot8

2. Write the value of each of the following :

(i) cos2

(ii) s i n 0 (iii)2cos3

(iv)3tan4

(v) sec 0 (vi) tan2

(vii)3tan2

(viii) c o s 2

16.2.1 Relation Between Trigonometric Functions

By definition x cos

y sin

As ytanx

, x 0

sincos ,

n2

andxcoty

, y 0

i.e.,cos 1cotsin tan n

Similarly,1sec

cosn2

Fig. 16.6

MATHEMATICS 71

Notes



and1cosec

sinn

Using Pythagoras theorem we have, 2 2x y 1

i.e., 2 2cos sin 1

or, 2 2cos sin 1

Note : 2cos is written as 2cos and 2sin as 2sin

Again 2 2x y 1

or2 2y 11 ,

x x for x 0 or, 2 21 tan sec

i.e. 2 2sec 1 tan

Similarly, 2 2cosec 1 cot

Example 16.9 Prove that 4 4 2 2sin cos 1 2sin cos

Solution : L.H.S. 4 4sin cos

4 4 2 2 2 2sin cos 2sin cos 2sin cos

22 2 2 2sin cos 2sin cos

2 2 2 21 2sin cos sin cos 1∵

= R.H.S.

Example 16.10 Prove that 1 sin

sec tan1 sin

Solution : L.H.S.1 sin1 sin

1 sin 1 sin1 sin 1 sin

2

21 sin1 sin

MATHEMATICS

Notes


72


2

21 sin

cos

1 sincos

1 sincos cos

sec tan R.H.S.

Example 16.11 If21sin29

, prove that 1sec tan 22

, given that lies in the first

quadrant.

Solution :21sin29

Also, 2 2sin cos 1

22 2 441 400 20cos 1 sin 1

841 841 29

20cos29

(cos is positive as lies in the first quardrant)

21tan20

29 21 29 21sec tan20 20 20

5 12 R.H.S.2 2


1. Prove that 4 4 2 2sin cos sin cos

2. If1tan2

, find the other five trigonometric functions.

3. Ifbcoseca

, find the other five trigonometric functions, if lies in the first quardrant.

4. Prove that 1 cos

cosec cot1 cos

MATHEMATICS 73

Notes



RealNumbers

5. If cot cosec 1.5 , show that 5cos

136. If tan sec m , find the value of cos

7. Prove that 2tan A 2 2tanA 1 5tanA 2sec A

8. Prove that 6 6 2 2sin cos 1 3sin cos

9. Prove that cos sin cos sin

1 tan 1 cot

10. Prove that tan sin cot cosec sec

1 cos 1 cos

16.3 TRIGONOMETRIC FUNCTIONS OF SOME SPECIFIC REAL NUMBERS

The values of the trigonometric functions of 0, 6

,4

,3

and 2

are summarised below in the

form of a table :

06 4 3 2

Function

sin 012

12

32

1

cos 13

212

12

0

tan 013 1 3 Not defined

As an aid to memory, we may think of the following pattern for above mentioned values of sinfunction :

04

,14

,24

,34

,44

On simplification, we get the values as given in the table. The values for cosines occur in thereverse order.Example 16.12 Find the value of the following :

(a) sin sin cos cos4 3 4 3

(b) 2 2 24tan cosec cos4 6 3

MATHEMATICS

Notes


74

Trigonometric Functions-ISolution :

(a) sin sin cos cos4 3 4 3

1 3 1 12 22 2

3 12 2

(b) 2 2 24tan cosec cos4 6 3

22 2 14 1 2

2

14 44

14

Example 16.13 If A3

and B6

, verify that

cos A B cosAcosB sinAsinB

Solution : L.H.S. cos A B

cos3 6

cos2

= 0

R.H.S. cos cos sin sin3 6 3 6

1 3 3 12 2 2 2

3 3 04 4

L.H.S. = 0 = R.H.S. cos (A + B) = cos A cos B sin A sin B

MATHEMATICS 75

Notes



CHECK YOUR PROGRESS 16.51. Find the value of

(i) 2 2 2sin tan tan6 4 3

(ii) 2 2 2 2sin cosec sec cos3 6 4 3

(iii)2 2cos cos sin sin3 3 3 3

(iv) 2 2 2 24cot cosec sec tan3 4 3 4

(v)1sin sin cos cos

6 4 3 4 4

2. Show that

2 21 tan tan tan tan sec sec6 3 6 3 6 3

3. Taking A , B3 6

, verify that

(i) tan A t a n Btan A B1 tan A tan B

(ii) cos A B cosAcosB s inAsinB

4. If4

, verify the following :

(i) sin2 2sin cos (ii) 2 2cos2 cos sin22cos 1

21 2sin

5. If A6

, verify that

(i) 2cos2A 2cos A 1 (ii) 22 t a n Atan2A

1 tan A

(iii) sin2A 2 s i n A c o s A

16.4 GRAPHS OF TRIGONOMETRIC FUNCTIONSGiven any function, a pictorial or a graphical representation makes a lasting impression on theminds of learners and viewers. The importance of the graph of functions stems from the fact thatthis is a convenient way of presenting many properties of the functions. By observing the graphwe can examine several characteristic properties of the functions such as (i) periodicity, (ii)intervals in which the function is increasing or decreasing (iii) symmetry about axes, (iv) maximumand minimum points of the graph in the given interval. It also helps to compute the areas enclosedby the curves of the graph.

MATHEMATICS

Notes


76


16.4.1 Variations of sin as Varies Continuously From 0 to 2Let X'OX and Y'OY be the axes ofcoordinates.With centre O and radiusOP = unity, draw a circle. Let OPstarting from OX and moving inanticlockwise direction make an angle

with the x-axis, i.e. XOP = . DrawPM X'OX, then sin MP as OP=1.

The variations of sin are the same as those ofMP.I Quadrant :

As increases continuously from 0 to 2

PM is positive and increases from 0 to 1. sin is positive.

II Quadrant ,2

In this interval, lies in the second quadrant.Therefore, point P is in the second quadrant. Here PM= y is positive, but decreases from 1 to 0 as varies

from to2

. Thus sin is positive.

III Quadrant 3

,2

In this interval, lies in the third quandrant. Therefore,point P can move in the third quadrant only. HencePM = y is negative and decreases from 0 to as

varies from to 32

. In this intervalsin decreases from 0 to 1. In this interval sin is

negative.

IV Quadrant3

, 22

In this interval, lies in the fourth quadrant. Therefore,point P can move in the fourth quadrant only. Here againPM = y is negative but increases from -1 to 0 as

varies from3 to 22

. Thus sin is negative in this

interval.

Fig. 16.7

Fig. 16.8

Fig. 16.9

Fig. 16.10

MATHEMATICS 77

Notes


Trigonometric Functions-I16.4.2 Graph of sin as varies from 0 to .Let X'OX and Y'OY be the two coordinate axes of reference. The values of are to be measuredalong x-axis and the values of sine are to be measured along y-axis.

(Approximate value of 1 32 1.41, .707, .8722

)

06 3 2

23

56

76

43

32

53

116 2

sin 0 .5 .87 1 .87 .5 0 .5 .87 1 .87 .5 0

Fig. 16.11

Some Observations (i) Maximum value of sin is 1.

(ii) Minimum value of sin is 1.(iii) It is continuous everywhere.

(iv) It is increasing from 0 to 2

and from 32

to 2 . It is decreasing from2

to 32

.

With the help of the graph drawn in Fig. 16.12 we can always draw another graph.y = sin in the interval of [2 , 4 ] ( see Fig. 16.11)1)

What do you observe ?

The graph of y = sin in the interval [2 , 4 ] is the same as that in 0 to2 . Therefore, thisgraph can be drawn by using the property sin (2 ) sin . Thus, sin repeats itself when

is increased by2 . This is known as the periodicity of sin .

Fig. 16.12

MATHEMATICS

Notes


78

Trigonometric Functions-IWe shall discuss in details the periodicity later in this lesson.Example 16.14 Draw the graph of y = sin 2 .

Solution :

: 06 4 3 2

23

34

56

2 : 03 2

23 3

32

53 2

sin2 : 0 .87 1 .87 0 .87 1 .87 0

Fig. 16.13

The graph is similar to that of y = sin

Some Observations1. The other graphs of sin , like a sin , 3 sin 2 can be drawn applying the same

method.

2. Graph of sin , in other intervals namely 4 , 6 , 2 , 0 , 4 , 2 ,can also be drawn easily. This can be done with the help of properties of allied

angles: sin 2 sin , sin 2 sin . i.e., repeats itself whenincreased or decreased by 2 .


1. What are the maximum and minimum values of sin in 0, 2 ?

2. Explain the symmetry in the graph of sin in 0, 2

3. Sketch the graph of y = 2 sin , in the interval 0,

MATHEMATICS 79

Notes



4. For what values of in , 2 , sin becomes

(a)1

2 (b)

32

5. Sketch the graph of y = sin x in the interval of ,

16.4.3 Graph of cos as Varies From 0 to 2

As in the case of sin , we shall also discuss the changes in the values of cos when assumes

values in the intervals 0 , ,2

, ,2

3,

2 and 3

, 2 .2

I Quadrant : In the interval 0 ,2

, point P lies in

the first quadrant, therefore, OM = x is positive but

decreases from 1 to 0 as increases from 0 to2

.

Thus in this interval cos decreases from 1 to 0.

cos is positive in this quadrant.

II Quadrant : In the interval ,2

, point P lies in

the second quadrant and therefore point M lies on thenegative side of x-axis. So in this case OM = x isnegative and decreases from 0 t o 1as increases

from2

to . Hence in this inverval cos decreases

from 0 to 1.

cos is negative.

III Quadrant : In the interval 3

,2 , point P lies

in the third quadrant and therefore, OM = x remainsnegative as it is on the negative side of x-axis. ThereforeOM = x is negative but increases from 1 to 0 as

increases from to 3 .2

Hence in this interval cos

increases from -1 to 0. cos is negative.

IV Quadrant : In the interval 3, 2

2, point P lies

Fig.16.14

Fig. 16.16

Fig. 16.15

MATHEMATICS

Notes


80

Trigonometric Functions-Iin the fourth quadrant and M moves on the positiveside of x-axis.Therefore OM = x is positive. Also it

increases from 0 to 1 as increases from 3to 2

2.

Thus in this interval cos increases from 0 to 1.

cos is positive.Let us tabulate the values of cosines of some suitablevalues of .

06 3 2

23

56

76

43

32

53

116 2

cos 1 .87 .5 0 0.5 .87 1 .87 .5 0 .5 .87 1

Fig. 16.18

Let X'OX and Y'OY be the axes. Values of are measured along x-axis and those of cosalong y-axis.

Some observations

(i) Maximum value of cos 1.

(ii) Minimum value of cos 1.

(iii) It is continuous everywhere.

(iv) cos 2 cos . Also 2 cos .Cos repeats itself when is

increased or decreased by2 . It is called periodicity of cos . We shall discuss indetails about this in the later part of this lesson.

(v) Graph of cos in the intervals [2 , 4 ] [4 , 6 ] [ 2 , 0] , will be the same asin [0, 2 ] .

Example 16.15 Draw the graph of cos as varies from to . From the graph readthe values of when cos 0.5 .

Fig. 16.17

MATHEMATICS 81

Notes


Trigonometric Functions-ISolution :

:56

23 2 3 6

06 3 2

23

56

cos : 1.0 0.87 0.5 0 .50 .87 1.0 0.87 0.5 0 .5 .87 1

cos 0.5

when ,3 3

cos 0.5

when 2 2,3 3

Fig. 16.19

Example 16.16 Draw the graph of cos 2 in the interval 0 to .

Solution :

06 4 3 2

23

34

56

2 03 2

23 3

32

53 2

c o s 2 1 0.5 0 0.5 1 0.5 0.5 1

Fig. 16.20

MATHEMATICS

Notes


82



1. (a) Sketch the graph of y = cos as varies from to4 4

.

(b) Draw the graph of y = 3 cos as varies from 0 to 2 .(c) Draw the graph of y = cos 3 from to and read the values of when

cos 0.87 and cos 0.87.

(d) Does the graph of y = cos in 3

,2 2

lie above x-axis or below x-axis?

(e) Draw the graph of y = cos in [2 , 4 ]

16.4.4 Graph of tan as Varies from 0 to

In I Quadrant : tan can be written as sincos

Behaviour of tan depends upon the behaviour of sin and1

cosIn I quadrant,sin increases from 0 to 1,cos decreases from 1 to 0

But1

cos increases from 1 indefintely (and write it as increasses from 1 to ) tan 0

tan increases from 0 to . (See the table and graph of tan ).

In II Quadrant :sintancos

sin decreases from 1 to 0.cos decreases from 0 to 1.tan is negative and increases from to 0

In III Quadrant :sintancos

sin decreases from 0 to 1cos increases from 1 t o 0tan is positive and increases from 0 to

In IV Quadrant :sintancos

sin increases from 1to0cos increases from 0 to 1tan is negative and increases form to0

Graph of tan

0 6 30

20

223

56

76

43

30

23

02

53

116 2

tan 0 .58 1.73 -1.73 -.58 0 .58 1.73 -1.73 -.58 0 0

MATHEMATICS 83

Notes



Fig. 16.21

Observations

(i) tan(180 ) tan . Therefore, the complete graph of tan consists ofinfinitely many repetitions of the same to the left as well as to the right.

(ii) Since tan( ) tan , therefore, if ,tan is any point on the graph then

( , tan ) will also be a point on the graph.(iii) By above results, it can be said that the graph of y tan is symmetrical in

opposite quadrants.(iv) tan may have any numerical value, positve or negative.

(v) The graph of tan is discontinuous (has a break) at the points 3,2 2

.

(vi) As passes through these values, tan suddenly changes from to .

16.4.5 Graph of cot as Varies From 0 to

The behaviour of cot depends upon the behaviour of cos and 1

sin as 1cot cossin

We discuss it in each quadrant.

I Quadrant :1cot cos

sin

cos decreases from 1 to 0

sin increases from 0 to 1

cot also decreases from to 0 but cot 0.

II Quadrant :1cot cos

sin

cos decreases from 0 t o 1

sin decreases from 1 to 0

MATHEMATICS

Notes


84


cot 0 or cot decreases from 0 to

III Quadrant :1cot cos

sin

cos increases from 1 t o 0

sin decreases from 0 to 1

cot decreases from to 0.

IV Quadrant :1cot cos

sin

cos increases from 0 to 1

sin increases from 1 t o 0

cot < 0

cot decreases from 0 to

Graph of cot

06 3 2

23

56 0 0

76

43

32

53

116 2

cot 1.73 .58 0 -.58 -1.73 1.73 .58 0 -.58 -1.73

Fig 16.22

Observations

(i) Since cot ( ) cot , the complete graph of cot consists of the portion from

0 to or 3to2 2

.

MATHEMATICS 85

Notes



Fig. 16.23

Fig. 16.24

Fig.16.25

(ii) cot can have any numerical value - positive or negative.

(iii) The graph of cot is discontinuous, i.e. it breaks at 0 , , 2 , .

(iv) As takes values 0, , 2 , cot suddently changes from to

CHECK YOUR PROGRESS 16.81. (a) What is the maximum value of tan ?

(b) What changes do you observe in 3tan at , ?

2 2

(c) Draw the graph of y tan from to . Find from the graph the value of forwhich tan 1.7.

2. (a) What is the maximum value of cot ?

(b) Find the value of whencot 1, from the graph.

16.4.6 To Find the Variations And Draw The Graph of sec As Varies From 0 to .

Let X'OX and Y'OY be the axes of coordinates. Withcentre O, draw a circle of unit radius.

Let P be any point on the circle. Join OP and drawPM X'OX.

OP 1secOM OM

Variations will depend upon OM.I Quadrant : sec is positive as OM is positive.

Also sec 0 = 1 and sec 2

when we approach 2

from the right.

As varies from 0 to 2

, sec increases from 1 to

.

II Quadrant : sec is negative as OM is negative.

sec2

when we approach 2

from the left. Also sec

1.

As varies from 2

to , sec changes from

MATHEMATICS

Notes


86


to 1.

It is observed that as passes through 2

, sec changes

from to .

III Quadrant : sec is negative as OM is negative.

sec 1 and 3sec2

when the angle approaches

32

in the counter clockwise direction. As varies from

to32

, sec decreases from 1to .

IV Quadrant : sec is positive as OM is positive. when

is slightly greater than 32

, sec is positive and very large.

Also sec 2 1. Hence sec decreases from to 1 as

varies from 3 to 22

.

It may be observed that as passes through

3 ; sec2

changes from to .

Graph of sec as varies from 0 to 2

0 6 30

20

223

56

76

43

30

23

02

53

116 2

cot 1 1.15 2 -2 -1.15 -1 -1.15 -2 2 1.15

1

Fig. 16.28

Fig.16.26

Fig.16.27

MATHEMATICS 87

Notes



Fig. 16.29

Fig. 16.30

Fig. 16.31

Observations

(a) sec cannot be numerically less than 1.

(b) Graph of sec is discontinuous, discontinuties (breaks) occuring at 2

and32

.

(c) As passes through 2

and32

, see changes abruptly from to and then

from to respectively..

16.4.7 Graph of cosec as Varies From 0 to

Let X'OX and Y'OY be the axes of coordinates. Withcentre O draw a circle of unit radius. Let P be anypoint on the circle. Join OP and draw PMperpendicular to X'OX.

OP 1cosecMP MP

The variation ofcosec will depend upon MP.

I Quadrant : cosec is positive as MP is positive.

cosec 12

when is very small, MP is also small and

therefore, the value of cosec is very large.

As varies from 0 to2

, cosec decreases from

to 1.

II Quadrant : PM is positive. Therefore, cosec is

positive. cosec 12

and cosec when the

revolving line approaches in the counter clockwisedirection.

As varies from2

to , cosec increases from

1 to .

III Quadrant :PM is negative

cosec is negative. When is slightly greater than ,

MATHEMATICS

Notes


88


cosec is very large and negative.

Also cosec 3 12

.

As varies from 3to2

, cosec changes from

to 1.

It may be observed that as passes through , cosec changes from to .

IV Quadrant :

PM is negative.

Therefore, cosec as approaches 2 .

as varies from 32

to 2 , cosec varies from

1 to .

Graph of cosec

06 3 2

23

56 0 0

76

43

32

53

116 2

cosec 2 1.15 1 1.15 2 2 1.15 1 1.15 2

Fig. 16.34

Observations

(a) cosec cannot be numerically less than 1.

(b) Graph of cosec is discountinous and it has breaks at 0, , 2 .

(c) As passes through , coses changes from to . The values at 0 and2 are and respectively..

Fig. 16.32

Fig. 16.33

MATHEMATICS 89

Notes



Example 16.17 Trace the changes in the values of sec as lies in to .Soluton :

Fig. 16.35

CHECK YOUR PROGRESS 16.91. (a) Trace the changes in the values of sec when lies between 2 and 2 and

draw the graph between these limits.(b) Trace the graph of cosec ,when lies between 2 and 2 .

16.5 PERIODICITY OF THE TRIGONOMETRIC FUNCTIONSFrom your daily experience you must have observed things repeating themselves after regularintervals of time. For example, days of a week are repeated regularly after 7 days and monthsof a year are repeated regularly after 12 months. Position of a particle on a moving wheel isanother example of the type. The property of repeated occurence of things over regular intervalsis known as periodicity.

Definition : A function f (x) is said to be periodic if its value is unchanged when the value of thevariable in increased by a constant, that is if f (x + p) = f (x) for all x.

If p is smallest positive constant of this type, then p is called the period of the function f (x).

If f (x) is a periodic function with period p, then 1

f x is also a periodic function with period p.

16.5.1 Periods of Trigonometric Functions

(i) sinx sin x 2n ; n = 0, 1, 2, .....

(ii) cosx cos x 2n ; n = 0, 1, 2 ,.....

Also there is no p, lying in 0 to 2 , for which

sinx sin x p

cosx cos x p , for all x

MATHEMATICS

Notes


90


2 is the smallest positive value for which

sin x 2 sinx and cos x 2 cosxsin x and cos x each have the period 2 .

(iii) The period of cosec x is also2 because cosec x 1sin x

.

(iv) The period of sec x is also 2 as sec x 1c o s x

.

(v) Also tan x t a n x.

Suppose p 0 p is the period of tan x, then

tan x p tanx, for all x.

Put x = 0, then tan p = 0, i.e., p = 0 or . the period of tan x is .

p can not values between 0 and for which tanx tan x p

The period of tanx is

(vi) Since 1cot xt a n x

, therefore, the period of cot x is also .

Example 16.18 Find the period of each the following functions :

(a) y = 3 sin 2x (b)xy cos2

(c) y =xtan4

Solution :

(a) Period is 22

, i.e., .

(b)1y cos x2

, therefore period 2 412

(c) Period of y = xtan 4144

CHECK YOUR PROGRESS 16.101. Find the period of each of the following functions :

(a) y = 2 sin 3x (b) y = 3 cos 2x

(c) y = tan 3x (d) 2y sin 2x

MATHEMATICS 91

Notes



WH

� An angle is generated by the rotation of a ray.� The angle can be negative or positive according as rotation of the ray is clockwise or

anticlockwise.� A degree is one of the measures of an angle and one complete rotation generates an

angle of 360°.� An angle can be measured in radians, 360° being equivalent to 2 radians.� If an arc of length l subtends an angle of radians at the centre of the circle with radius

r, we have l = r .� If the coordinates of a point P of a unit circle are (x, y) then the six trigonometric functions

are defined as sin y , cos x , ytanx

, xcoty

, 1secx

and

1cosecy .

The coordinates (x, y) of a point P can also be written as cos , sin .

Here is the angle which the line joining centre to the point P makes with the positivedirection of x-axis.

� The values of the trigonometric functions sin and cos when takes values 0,

, , ,6 4 3 2

are given by

06 4 3 2

sin 012

12

32

1

cos 13

212

12

0

� Graphs of sin , cos are continous every where

— Maximum value of both sin and cos is 1.

— Minimum value of both sin and cos is -1.— Period of these functions is 2 .

LET US SUM UP

MATHEMATICS

Notes


92


� tan and cot can have any value between and .

— The function tan has discontinuities (breaks) at 2

and 32

in 0, 2 .

— Its period is .— The graph of cot has discontinuities (breaks) at 0, , 2 . Its period is .

� sec cannot have any value numerically less than 1.

(i) It has breaks at 2

and 32

. It repeats itself after2 .

(ii) cosec cannot have any value between 1 and +1.It has discontinuities (breaks) at 0, , 2 .It repeats itself after 2 .

� http://www.wikipedia.org� http://mathworld.wolfram.com

TERMINAL EXERCISE ONS

1. A train is moving at the rate of 75 km/hour along a circular path of radius 2500 m.Through how many radians does it turn in one minute ?

2. Find the number of degrees subtended at the centre of the circle by an arc whose lengthis 0.357 times the radius.

3. The minute hand of a clock is 30 cm long. Find the distance covered by the tip of the handin 15 minutes.

4. Prove that

(a)1 sin sec tan1 sin

(b)1 sec tan

sec tan

(c) 2 2tan cot 2sin cos

1 tan 1 cot(d) 21 sin tan sec

1 sin

(e) 8 8 2 2 2 2sin cos sin cos 1 2sin cos

(f) 2 2sec cosec tan cot

5. If4

, verify that 3sin3 3sin 4sin

SUPPORTIVE WEB SITES

MATHEMATICS 93

Notes


Trigonometric Functions-I6. Evaluate :

(a)25sin

6(b)

21sin4

(c) 3tan

4

(d)17sin4

(e)19cos3

7. Draw the graph of cos x from x2

to 3x2

.

8. Define a periodic function of x and show graphically that the period of tan x is , i.e. theposition of the graph from x = to 2 is repetition of the portion from x = 0 to .

MATHEMATICS

Notes


94


ANSWERS


1. (i)3

(ii)12

(iii)512

(iv)712

(v)32

2. (i) 45° (ii) 15° (iii) 9° (iv) 3° (v) 120°

3.4

,1336

,1436

4.56

5.3


1. (a)6

(b)3

(c)56

2. (a) 36° (b) 30° (c) 20°

3.16

radian; 9.55° 4.15

radian 5. 95.54 m

6. (a) 0.53 m (b) 38.22 cm (c) 0.002 radian(d) 12.56 m (e) 31.4 cm (f) 3.75 radian(g) 6.28 m (h) 2 radian (i) 19.11 m.

CHECK YOUR PROGRESS 16.31. (i) ive (ii) ive (iii) ive (iv) + ive

(v) + ive (vi) ive (vii) + ive (viii) ive

2. (i) zero (ii) zero (iii)12

(iv) 1

(v) 1 (vi) Not defined (vii) Not defined (viii) 1


2.1

sin5 ,

2cos

5, cot 2 , cosec 5 ,

5sec2

3.asinb

,2 2b acosb

, 2 2

bsecb a

,

2 2

atanb a

,2 2b acota

6. 22m

1 m

MATHEMATICS 95

Notes




1. (i)144

(ii)162

(iii) 1 (iv)223

(v) Zero

CHECK YOUR PROGRESS 16.61. 1, 1 3. Graph of y = 2 sin , 0,

Fig. 16.36

4. (a)7 11,6 6

(b)4 5,3 3

5. y sinxfrom to

Fig. 16.37


1. (a) y cos , to4 4

Fig. 16.38

MATHEMATICS

Notes


96

Trigonometric Functions-I(b) y 3cos ; 0 to 2

Fig. 16.39

(c) y cos3 , to

cos 0.87

,6 6

cos 0.87

5 5,6 6

(d) Graph of y cos in 3

,2 2 lies below the x-axis.

(e) y cos

lies in 2 to 4

Fig. 16.41


1. (a) Infinite (b) At 2

,32

there are breaks in graphs.

Fig. 16.40

MATHEMATICS 97

Notes



(c) y tan2 , to

At ,tan 1.73

2. (a) Infinite (b)3cot 1 at4

CHECK YOUR PROGRESS 16.91. (a) y sec

Fig. 16.42

Points of discontinuity of sec2 are at 3,

4 4 in the interval 0, 2 .

(b) In tracing the graph from 0 to 2 , use cosec cosec .


1. (a) Period is 23

(b) Period is 22

(c) Period of y is3

(d) 2 1 cos4x 1 1y sin 2x cos4x2 2 2

; Period of y is 2 i.e4 2

(e)x 1

y 3cot3 , Period of y is 31

3TERMINAL EXERCISE

1.1 radian2

2. 20.45° 3. 15 cm

6. (a)12

(b)12 (c) 1 (d)

12 (e)

12

MATHEMATICS

Notes


98


7.

Fig. 16.43

8.

y secFig. 16.44

MATHEMATICS 99

Notes


Trigonometric Functions-II

17

TRIGONOMETRIC FUNCTIONS-II

In the previous lesson, you have learnt trigonometric functions of real numbers, draw and interpretthe graphs of trigonometric functions. In this lesson we will establish addition and subtractionformulae for cos A B , sin A B and tan A B . We will also state the formulae forthe multiple and sub multiples of angles and solve examples thereof. The general solutions ofsimple trigonometric functions also discussed in the lesson.


write trigonometric functions of xx, , x y, x, x2 2

where x, y are real nunbers;

establish the addition and subtraction formulae for :

cos (A B) = cos A cos B ∓ sin A sin B,

sin (A B) = sin A cos B cos A sin B and tanA tanBtan A B

1 tanAtanB∓

solve problems using the addition and subtraction formulae;state the formulae for the multiples and sub-multiples of angles such as cos2A, sin 2A, tan

2A, cos 3A, sin 3A, tan 3A, A Asin ,cos2 2

and Atan2

; and

solve simple trigonometric equations of the type :

sinx 0,cosx 0 , tanx 0,

sinx sin , cosx cos , tanx tan

MATHEMATICS

Notes


100


EXPECTED BACKGROUND KNOWLEDGEDefinition of trigonometric functions.Trigonometric functions of complementary and supplementary angles.Trigonometric identities.

17.1 ADDITION AND MULTIPLICATION OF TRIGONOMETRIC FUNCTIONSIn earlier sections we have learnt about circular measure of angles, trigonometric functions,values of trigonometric functions of specific numbers and of allied numbers.

You may now be interested to know whether with the given values of trigonometric functions ofany two numbers A and B, it is possible to find trigonometric functions of sums or differences.

You will see how trigonometric functions of sum or difference of numbers are connected withthose of individual numbers. This will help you, for instance, to find the value of trigonometric

functions of 12

and 512

etc.

12 can be expressed as

4 6

512

can be expressed as 4 6

How can we express 712

in the form of addition or subtraction?

In this section we propose to study such type of trigonometric functions.17.1.1 Addition Formulae

For any two numbers A and B,


In given figure trace out

SOP A

POQ B

SOR B

where points P, Q, R, S lie on the unit circle.Coordinates of P, Q, R, S will be (cos A, sin A), [cos (A + B), sin (A + B)],

cos B ,sin B , and (1, 0).

From the given figure, we have

Fig. 17.1

MATHEMATICS 101

Notes



side OP = side OQ

POR = QOS (each angle = B + QOR)

side OR = side OS

POR QOS (by SAS)

PR = QS

22PR cosA cosB sinA sin B

2 2QS cos A B 1 sin A B 0

Since 2 2PR QS

2 2 2 2cos A cos B 2cosAcosB sin A sin B 2sinAsinB

2 2cos A B 1 2cos A B sin A B

1 1 2 cosAcosB sinAsinB 1 1 2cos A B

cosAcosB sinAsinB cos A B (I)

Corollary 1For any two numbers A and B, cos (A B) = cos A cos B + sin A sin BProof : Replace B by B in (I)


[∵ cos B cosB and sin B sinB]

Corollary 2For any two numbers A and B

sin A B sinAcosB cosAsinB

Proof : We know that cos A s i n A2

and sin A cosA2

sin A B cos A B2

cos A B2

Fig. 17.2

MATHEMATICS

Notes


102


cos A cosB sin A sinB2 2

or sin A B sinAcosB cosAsinB .....(II)

Corollary 3For any two numbers A and B

sin A B sinAcosB cosAsinB

Proof : Replacing B by B in (2), we have

sin A B sinAcos B cosAsin B

or sin A B sinAcosB cosAsinB

Example 17.1

(a) Find the value of each of the following :

(i) sin12

(ii) cos12

(iii)7cos12

(b) If1sinA ,10

1sinB5 show that A B

4Solution :

(a) (i)5sin sin sin cos cos sin12 4 6 4 6 4 6

1 3 1 1 3 12 22 2 2 2

5 3 1sin12 2 2

(ii) cos cos12 4 6

cos cos sin sin4 6 4 6

1 3 1 1 3 12 22 2 2 2

3 1cos12 2 2

Observe that5sin cos12 12

MATHEMATICS 103

Notes



(iii)7cos cos12 3 4

cos cos sin sin3 4 3 4

1 1 3 1 1 32 22 2 2 2

7 1 3cos12 2 2

(b) sin A B sinAcosB cosAsinB

1 3cosA 110 10

and 1 2cosB 15 5

Substituting all these values in (II), we get

1 2 3 1sin A B10 5 10 5

5 5 5 1 sin410 5 50 5 2 2

or A B4


1. (a) Find the values of each of the following :

(i) sin12

(ii)2 2sin cos cos sin

9 9 9 9(b) Prove the following :

(i)1sin A cosA 3 sinA

6 2 (ii) 1sin A cosA sinA

4 2

(c) If 8s inA17

and 5sinB13

, find sin A B

2. (a) Find the value of 5cos .12

(b) Prove the following :

(i) cos sin 2cos4

(ii) 3 sin cos 2sin6

MATHEMATICS

Notes


104


(iii) cos n 1 A cos n 1 A sin n 1 Asin n 1 A cos2A

(iv) cos A cos B sin A sin B cos A B4 4 4 4

Corollary 4 : tanA tanBtan A B1 tanAtanB

Proof : sin A Btan A Bcos A B

sin A cosB cosAsinBcosAcosB sin A sinB

Dividing by cos A, cos B, we have

sinAcosB cosAsinBcosAcosB cosAcosBtan A B cosAcosB sinAsinBcosAcosB cosAcosB

or tanA tanBtan A B1 tanAtanB

......(III)

Corollary 5 : tanA tanBtan A B1 tanAtanB

Proof : Replacing B by B in (III), we get the required result.

Corollary 6 : cotAcotB 1cot A BcotB cotA

Proof : cos A B cosAcosB sinAsinBcot A Bsin A B sinAcosB cosAsinB

Dividing by sin A sin B, we have ......(IV)cotAcotB 1cot A BcotB cotA

Corollary 7 : 1 tanAtan A4 1 tanA

Proof : tan tanA

4tan A4 1 tan tanA

4

1 tanA1 tanA

as tan 14

MATHEMATICS 105

Notes



Similarly, it can be proved that

1 tanAtan A4 1 tanA

Example 17.2 Find tan12

Solution : tan tan

4 6tan tan12 4 6 1 tan tan

4 611 3 131 3 11 1.3

3 1 3 1 4 2 323 1 3 1

2 3 tan 2 312

Example 17.3 Prove the following :

(a)

7 7cos sin 436 36 tan7 7 9cos sin36 36

(b) tan 7 A tan 4 A tan3A tan7A tan4A tan3A

(c) 7 5tan tan 2tan18 9 18

Solution : (a) Dividing numerator and denominator by 7cos36

, we get

L.H.S.

7 7 7cos sin 1 tan36 36 367 7 7cos sin 1 tan36 36 36

7tan tan4 36

71 tan tan4 36

7tan4 36

MATHEMATICS

Notes


106

Trigonometric Functions-II16 4tan tan R.H.S.36 9

(b) tan4A tan3Atan 7 A tan 4A 3A1 tan4A tan3A

or tan 7 A tan7A tan 4 A tan 3A tan4A tan3A

or tan7A tan4A tan3A tan7Atan4Atan3A

(c)

5 2tan tan7 5 2 18 18tan tan 5 218 18 18 1 tan tan18 18

7 7 5 2 5 2tan tan tan tan tan tan18 18 18 18 18 18

.....(1)

7 2tan tan cot cot18 2 9 9 18

(1) can be written as

7 2 5 2 5tan cot tan tan tan tan18 18 18 18 18 9

7 5tan tan 2tan18 9 18


1. Fill in the blanks :

(i) sin A sin A .........4 4

(ii) cos cos .........3 4 3 4

2. (a) Prove the following :

(i) tan tan 1.4 4

(ii)cotAcotB 1cot A BcotB cot A

(iii) tan tan tan tan 112 6 12 6

(b) If atan Ab

;ctan Bd

, Prove that

ad bctan A B .bd ac

MATHEMATICS 107

Notes



(c) Find the value of 11cos12

.

3. (a) Prove the following :

(i) 3tan A tan A 1

4 4 (ii)

cos sintan

cos sin 4

(iii)cos sin

tancos sin 4

17.2 TRANSFORMATION OF PRODUCTS INTO SUMS AND VICE VERSA

17.2.1 Transformation of Products into Sums or DifferencesWe know that

sin A B sin A c o s B cosAsinB

sin A B sinAcosB cosAs inB


cos A B cosAcosB s inAs inB

By adding and subtracting the first two formulae, we get respectively

2sinAcosB sin A B sin A B .....(1)

and 2cosAsinB sin A B sin A B .....(2)

Similarly, by adding and subtracting the other two formulae, we get

2cosAcosB cos A B cos A B ....(3)

and 2sinAsinB cos A B cos A B ....(4)

We can also quote these as

2s inAcosB sin sum sin difference

2cosAsinB sin sum sin difference

2cosAcosB cos sum cos difference

2sinAsinB cos difference cos sum

17.2.2 Transformation of Sums or Differences into Products

In the above results put

MATHEMATICS

Notes


108

Trigonometric Functions-IIA + B = C

and A B = D

Then C DA2

and C DB2

and (1), (2), (3) and (4) become

C D C DsinC sin D 2sin cos2 2

C D C DsinC sin D 2cos sin2 2

C D C DcosC cosD 2cos cos2 2

C D C DcosD cosC 2sin sin2 2

17.2.3 Further Applications of Addition and Subtraction FormulaeWe shall prove that

(i) 2 2sin A B sin A B sin A sin B

(ii) 2 2cos A B cos A B cos A sin B or 2 2cos B sin A

Proof : (i) sin A B sin A B

sin A c o s B cosAsinB sinAcosB cosAsinB

2 2 2 2sin Acos B cos Asin B2 2 2 2sin A 1 sin B 1 sin A sin B

2 2sin A sin B

(ii) cos A B cos A B

cosAcosB sin A s i n B cosAcosB sinAsinB

2 2 2 2cos Acos B sin Asin B

2 2 2 2cos A 1 sin B 1 cos A sin B

2 2cos A sin B

2 21 sin A 1 cos B

2 2cos B sin A

Example 17.4 Express the following products as a sum or difference

(i) 2sin3 cos2 (ii) cos6 cos (iii)5sin sin12 12

MATHEMATICS 109

Notes


Trigonometric Functions-IISolution :

(i) 2sin3 cos2 sin 3 2 sin 3 2

sin5 sin

(ii)1cos6 cos 2cos6 cos2

1 cos 6 cos 62

1 cos7 cos52

(iii)5 1 5

sin sin 2sin sin12 12 2 12 12

1 5 5cos cos

2 12 12

1cos cos

2 3 2

Example 17.5 Express the following sums as products.

(i)5 7cos cos9 9

(ii)5 7sin cos36 36

Solution :

(i)5 7 5 7 5 7cos cos 2cos cos9 9 9 2 9 2

22cos cos3 9

cos cos9 9

∵

2cos cos3 9

2cos cos3 9

cos9

1cos

3 2∵

(ii)5 7 13 7

sin cos sin cos36 36 2 36 36

13 7cos cos36 36

MATHEMATICS

Notes


110


13 7 13 72cos cos36 2 36 2

52cos cos18 12

Example 17.6 Prove that cos7A cos9A tan8Asin9A sin7A

Solution :

L.H.S.

7A 9A 9A 7A2sin sin2 2

9A 7A 9A 7A2cos sin2 2

sin8A sinA sin8A tan8A R.H.S.cos8A sin A cos8A


(i) 2cos A sin B sin A B cos A B4 4

(ii) 2A A 1sin sin sinA

8 2 8 2 2Solution :

(i) Applying the formula2 2cos A sin B cos A B cos A B , we have

L.H.S. cos A B cos A B4 4 4 4

cos A B cos A B2

sin A B cos A B R.H.S.(ii) Applying the formula

2 2sin A sin B sin A B sin A B , we have

L.H.S.A A A A

sin sin8 2 8 2 8 2 8 2

sin sin A4

1sinA R.H.S.

2

MATHEMATICS 111

Notes




2 4 1cos cos cos cos9 9 3 9 16

Solution : L.H.S.2 4

cos cos cos cos3 9 9 9

1 1 2 42cos cos cos

2 2 9 9 91

cos3 2

∵

1 4cos cos cos

4 3 9 9

1 4 1 4cos 2cos cos

8 9 8 9 9

1 4 1 5cos cos cos

8 9 8 9 3

1 4 1 5 1cos cos8 9 8 9 16

.....(1)

Now5 4 4

cos cos cos9 9 9 .....(2)

From (1) and (2), we get

L.H.S.1 R.H.S.

16


1. Express each of the following as sums or differences :

(a) 2cos3 s i n 2 (b) 2sin4 sin2

(c) 2cos cos4 12

(d) 2sin cos3 6

2. Express each of the following as a product :

(a) sin6 s in4 (b) sin7 sin3

(c) cos2 cos4 (d) cos7 cos5

MATHEMATICS

Notes


112


3. Prove the following :

(a)5 4sin cos cos18 9 9

(b)

7cos cos9 18 17sin sin18 9

(c)5 7sin sin sin 018 18 18

(d)5 7cos cos cos 0

9 9 94. Prove the following :

(a) 2 2sin n 1 sin n sin 2n 1 sin

(b) 2 2cos cos 2 cos sin

(c) 2 2 3cos sin4 12 4

5. Show that 2 2cos sin4 4 is independent of .


(a)sin sin3 sin5 sin7 tan 4

cos cos3 cos5 cos 7

(b)5 5 7 1sin sin sin sin

18 6 18 18 16

(c) 2 2 2cos cos sin sin 4cos2

17.3 TRIGONOMETRIC FUNCTIONS OFMULTIPLES OF ANGLES

(a) To express sin 2A in terms of sin A, cos A and tan A.We know that

sin A B sin A c o s B cosAsinBBy putting B = A, we get

sin2A sinAcosA cosAsinA2sinAcosA

sin 2A can also be written as

2 22s inAcosAsin2A

cos A sin A(∵ 2 21 cos A sin A )

Dividing numerator and denomunator by 2cos A , we get

MATHEMATICS 113

Notes



22 2

2 2

sin A c o s A2cos Asin2A

cos A sin Acos A cos A

22 tan A

1 tan A

(b) To express cos 2A in terms of sin A, cos A and tan A.We know that

cos A B cosAcosB s inAs inB

Putting B = A, we have

cos2A cosAcosA sin A s i n A

or 2 2cos2A cos A sin A

Also 2 2cos2A cos A 1 cos A

2 2cos A 1 cos A

i.e, 2cos2A 2cos A 1 2 1 cos2Acos A2

Also 2 2cos2A cos A sin A2 21 sin A sin A

i.e., 2cos2A 1 2sin A 2 1 cos2Asin A2

2 2

2 2cos A sin Acos2Acos A sin A

Dividing the numerator and denominator of R.H.S. by 2cos A , we have

2

21 tan Acos2A1 tan A

(c) To express tan 2A in terms of tan A.

tanA tanAtan2A tan A A1 tanAtanA

22 t a n A

1 tan AThus we have derived the following formulae :

22 t a n Asin2A 2s inAcosA

1 tan A2

2 2 2 22

1 tan Acos2A cos A sin A 2cos A 1 1 2sin A1 tan A

MATHEMATICS

Notes


114


22 tanAtan2A

1 tan A

2 1 cos2Acos A2

, 2 1 cos2Asin A2

Example 17.9 If A6

, verify the following :

(i) 22 t a n Asin2A 2s inAcosA

1 tan A

(ii)2

2 2 22

1 tan Acos2A cos A sin2A 2cos A 1 1 2sin A1 tan A

(iii) 22 t a n Atan2A

1 tan ASolution :

(i)3sin2A sin

3 2

1 3 32s inAcosA 2sin cos 26 6 2 2 2

2 2

122tan2tanA 2 3 33611 tan A 4 231 tan 1

6 3Thus, it is verified that

22 t a n Asin2A 2s inAcosA

1 tan A

(ii)1cos2A cos

3 22 2

2 2 2 2 3 1 3 1 1cos A sin A cos sin6 6 2 2 4 4 2

22 2 32cos A 1 2cos 1 2 1

6 2

3 12 14 2

22 1 1 11 2sin A 1 2sin 1 2 1 2

6 2 4 2

MATHEMATICS 115

Notes



222

22 2

1 111 tan 11 tan A 2 3 136 311 tan A 3 4 211 tan 116 33

Thus, it is verified that2

2 2 2 22


(iii) tan 2 A tan 33

2 2

122tan2tanA 2 336 3.11 tan A 231 tan 16 3

Thus, it is verified that 22 t a n Atan2A

1 tan A

Example 17.10 Prove that sin2A tanA

1 cos2A

Solution : 2sin2A 2s inAcosA

1 cos2A 2cos A

s i n AcosA

= tan A

Example 17.11 Prove that cot A tan A 2cot2A.

Solution : 1cotA tan A tan AtanA

21 tan Atan A

22 1 tan A2 t a n A

2

22 t a n A

1 tan A

2tan2A 2cot2A.

Example 17.12 Evaluate 2 2 3cos cos .8 8

MATHEMATICS

Notes


116


Solution : 2cos8

2

31 cos 1 cos3 4 4cos8 2 2

1 12 2

2 2

2 1 2 12 2 1

Example 17.13 Prove that cosA A

tan .1 sinA 4 2

Solution : R.H.S. =

Atan tanA 4 2tan A4 2 1 tan tan4 2

Asin21 A A Acos cos sin2 2 2A A Asin cos sin2 2 21 Acos2

2

A A A Acos sin cos sin2 2 2 2

A Acos sin2 2

[Multiplying Numerator and Denominator by cosA sinA

2 2

2 2

2 2

A Acos sin2 2

A A A Acos sin 2cos sin2 2 2 2

cosA L.H.S1 s inA

.


2 2 2cos cos sin sin 4sin2

MATHEMATICS 117

Notes



Solution : 2 2cos cos sin sin

2 2 2 2cos cos 2cos cos sin sin 2sin sin

2 2 cos cos in sin

2 1 cos

2 22 2sin 4sin2 2


1. If A3

, verify that

(a) 22 t a n Asin2A 2s inAcosA

1 tan A

(b)2

2 2 2 22


2. Find the value of sin 2A when (assuming 0 < A < 2

)

(a)3cosA5

(b)12s inA13

(c)16tan A63

.

3. Find the value of cos 2A when

(a)15cosA17

(b)4s inA5

(c)5tan A

124. Find the value of tan 2A when

(a)3tan A4

(b)atan Ab

5. Evaluate 2 2 3sin sin .8 8


(a) 21 sin2Atan A

1 sin2A 4 (b)2

2cot A 1 sec2Acos A 1

7. (a) Prove that sin2A cosA

1 cos2A (b) Prove that tan A cotA 2cosec2A .

8. (a) Prove thatcosA A

tan1 sinA 4 2

(b) Prove that 2 2 2cos cos sin sin 4cos2

MATHEMATICS

Notes


118


17.3.1 Trigonometric Functions of 3A in Terms of those of A(a) sin 3A in terms of sin A

Substituting 2A for B in the formula

sin A B sin A c o s B cosAsinB, we get

sin A 2A sinAcos2A cosAsin2A

2sinA 1 2sin A cosA 2sinAcosA

3 2sinA 2sin A 2sinA 1 sin A

3 3sinA 2sin A 2sinA 2sin A

3sin3A 3sinA 4sin A ....(1)

(b) cos 3A in terms of cos ASubstituting 2A for B in the formula

cos A B cosAcosB s inAs inB, we get

cos A 2A cosAcos2A sinAsin2A

2cosA 2cos A 1 sin A 2sinAcosA

3 22cos A cosA 2cosA 1 cos A

3 32cos A cosA 2cosA 2cos A3cos3A 4cos A 3cosA ....(2)

(c) tan 3A in terms of tan APutting B = 2A in the formula

tanA tan Btan A B1 t a n A t a n B

, we get

tanA tan2Atan A 2A1 tanAtan2A

2

2

2 t a n AtanA1 tan A

2 tanA1 tan A1 tan A

3

22 2

2

tan A tan A 2tanA1 tan A

1 tan A 2tan A1 tan A

MATHEMATICS 119

Notes



3

23tanA tan A

1 3tan A....(3)

(d) Formulae for 3sin A and 3cos A

∵ 3sin3A 3sinA 4sin A

34sin A 3sinA sin3A

or 3 3sinA sin 3Asin A4

Similarly, 3cos3A 4cos A 3cosA

33cosA cos3A 4cos A

or 3 3cosA cos3Acos A4

Thus, we have derived the following formulae :3sin3A 3sinA 4sin A

3cos3A 4cos A 3cosA

3

23tanA tan Atan3A

1 3tan A

3 3sinA sin 3Asin A4

3 3cosA cos3Acos A4

Example 17.15 If A4

, verify that

(i) 3sin3A 3sinA 4sin A (ii) 3cos3A 4cos A 3cosA

(iii)3

23tanA tan Atan3A

1 3tan ASolution :

(i)3 1

sin3A sin4 2

3 33sinA 4sin A 3sin 4sin4 4

31 13 42 2

MATHEMATICS

Notes


120


3 4 12 2 2 2

Thus, it is verified that 3sin3A 3sinA 4sin A

(ii) 3 1cos3A cos

4 2

33 1 14cos A 3cosA 4 3

2 2

4 3 12 2 2 2

Thus, it is verified that 3cos3A 4cos A 3cosA

(iii)3tan3A tan 14

,

3 3

2 23tanA tan A 3 1 1 2 1

1 3tan A 1 3 1 2

Thus, it is verified that 3

23tanA tan Atan3A

1 3tan A

Example 17.16 If 1 1

cos a2 a

, prove that 33

1 1cos3 a

2 a

Solution : 3cos3 4cos 3cos

31 1 1 14 a 3 a2 a 2 a

3 22 3

1 1 1 1 3a 34 a 3a 3a8 a 2 2aa a

33

3 3a 1 1 1a2 22a a


1sin sin sin s in3

3 3 4

Solution : sin sin sin3 3

1 2sin cos2 cos

2 3

MATHEMATICS 121

Notes



2 21 sin 1 2sin 1 2sin2 3

2 212 sin sin sin

2 3

323 3sin 4sin 1

sin sin s in34 4 4


3 3 3cos Asin3A sin Acos3A sin4A4

Solution : 3 3cos Asin3A sin Acos3A3 3 3 3cos A 3sinA 4sin A sin A 4cos A 3cosA

3 3 3 3 3 33sinAcos A 4sin Acos A 4sin Acos A 3sin AcosA

3 33sinAcos A 3sin A c o s A

2 23sinAcosA cos A sin A 3sinAcosA cos2A

3sin2A cos2A2

3 sin4A 3 sin4A2 2 4

.

Example 17.19 Prove that 3 3 3cos sin cos sin

9 18 4 9 18

Solution : L.H.S.1 1

3cos cos 3sin sin4 9 3 4 18 6

3 1 1 1cos sin

4 9 18 4 2 2

3cos sin R.H.S.

4 9 18


1. If A3

, verify that

(a) 3sin3A 3sinA 4sin A (b) 3cos3A 4cos A 3cosA

MATHEMATICS

Notes


122


(c)3

23tanA tan Atan3A

1 3tan A

2. Find the value of sin 3A when (a) 2s inA3

(b) psinA .q

3. Find the value of cos 3A when (a) 1cosA3

(b) ccosA .d

4. Prove that 1

cos cos cos cos3 .3 3 4

5. (a) Prove that 3 32 3 2sin sin sin sin

9 9 4 9 9

(b) Prove that sin3A cos3Asin A cosA

is constant.

6. (a) Prove that 3

2cot A 3co tAcot3A

3cot A 1

(b) Prove that

3cos10A cos8A 3cos4A 3cos2A 8cosAcos 3A

17.4 TRIGONOMETRIC FUNCTIONS OFSUBMULTIPLES OF ANGLES

A A A, ,2 3 4

are called submultiples of A.

It has been proved that

2 1 cos2Asin A2

, 2 1 cos2Acos A2

, 2 1 cos2Atan A1 cos2A

Replacing A by A2

, we easily get the following formulae for the sub-multiple A2

:

A 1 cosAsin

2 2,

A 1 cosAcos2 2

andA 1 cosAtan2 1 cosA

We will choose either the positive or the negative sign depending on whether corresponding

value of the function is positive or negative for the value of A2

. This will be clear from the

following examples

MATHEMATICS 123

Notes



Example 17.20 Find the values of cos12

and cos24

.

Solution : We use the formulae A 1 cosAcos2 2

and take the positive sign, because

cos12

and cos24

are both positive.

1 cos6cos

12 2

312

2

2 32 2

4 2 38

23 1

8

24 2 3 1 3 2 3 1 3∵

3 12 2

1 cos12cos

24 2

3 112 22

2 2 3 14 2

4 6 28

Example 17.21 Find the values of sin8

and cos8

.

Solution : We use the formulaA 1 cosAsin2 2

and take the lower sign, i.e., negative sign, because sin8

is negative.

MATHEMATICS

Notes


124


1 cos4sin

8 2

112

2

2 1 2 222 2

Similarly,1 cos

4cos8 2

112

2

2 12 2

2 24

2 22

Example 17.22 If 7cosA25

and 3 A 22

, find the values of

(i)Asin2

(ii)Acos2

(iii)Atan2

Solution : ∵ A lies in the 4th-quardrant, 3 A 22

A34 2

Asin 02

,Acos 02

,Atan 0.2

7251A 1 cosA 18 9 3sin

2 2 2 50 25 5

MATHEMATICS 125

Notes



7251A 1 cosA 32 16 4cos

2 2 2 50 25 5

and725725

1A 1 cosA 18 9 3tan2 1 cosA 32 16 41


1. If A3

, verify that

(a)A 1 cosAsin2 2

(b)A 1 cosAcos2 2

(c)A 1 cosAtan2 1 cosA

2. Find the values of sin12

and sin .24

3. Determine the values of

(a) sin8

(b) cos8

(c) tan8

.

Example 17.23 Prove that following :

(a)5 1sin

10 4 and 10 2 5cos

10 4

(b)5 1cos

5 4 and 10 2 5sin

5 4

Solution : (a) Let A10

5A2

2A 3A2

sin2A sin 3A cos3A2

32s inAcosA 4cos A 3cosA

or 2cosA 2sinA 4cos A 3 0 .....(1)

MATHEMATICS

Notes


126


As cosA 0 and 2 21 sin A cos A

(1) becomes 22sinA 4 1 sin A 3 0

24sin A 2sinA 1 0

2 4 16 1 5s inA8 4

5 1s inA4

5 1sin10 4

Now2

2 5 1 10 2 5cos 1 sin 110 10 4 4

(b) Let A10

, 2A5

2cos2A 1 2sin A2

5 1 6 2 5 2 2 5cos 1 2 1 25 4 16 8

5 14

Now 2 10 2 5sin 1 cos5 5 4


tan 2tan2 4tan4 8cot8 cot

Solution : We have to prove that

tan cot 2tan2 4tan4 8cot8 0

orsin cos

2 tan 2 4tan4 8cot8 0cos sin

or2 c o s 2 2 tan2 4 tan 4 8cot8 0

2sin cos

orcos22 2tan2 4tan4 8cot8 0s i n 2

MATHEMATICS 127

Notes



or 2cot2 2tan2 4tan4 8cot8 0 ......(1)

Combining 2cot2 2 tan2

(1) becomes 4cot4 4tan4 8cot8 0 ......(2)

Combining 4cot4 4 tan4 ; L.H.S. of (2) becomes

8cot8 8cot8 0 R.H.S. of (2)

17.5 TRIGONOMETRIC EQUATIONS

You are familiar with the equations like simple linear equations, quadratic equations in algebra.You have also learnt how to solve the same.Thus, (i) x 3 0 gives one value of x as a solution.

(ii) 2x 9 0 gives two values of x.

You must have noticed, the number of values depends upon the degree of the equation.Now we need to consider as to what will happen in case x 's and y's are replaced by trigonometricfunctions.

Thus solution of the equation sin 1 0 , will give

sin 1 and 5 9, , ,....2 2 2

Clearly, the solution of simple equations with only finite number of values does not necessarilyhold good in case of trigonometric equations.So, we will try to find the ways of finding solutions of such equations.

17.5.1 To find the general solution of the equation sin = 0It is given that sin 0But we know that sin 0, sin , sin2 ,.. . . ,sinn are equal to 0

n , n NBut we know sin sin 0

sin , sin 2 , sin 3 ,....,sin n 0

n , n I .Thus, the general solution of equations of the type sin = 0 is given by = n where n is aninteger.

17.5.2 To find the general solution of the equation cos = 0It is given that cos 0

But in practice we know that cos 02

. Therefore, the first value of is

2.....(1)

MATHEMATICS

Notes


128


We know that cos cos or cos cos 0.2 2

or3cos 02

In the same way, it can be found that

5 cos7 9cos , , cos ,.....,2 2 2

cos 2n 12

are all zero.

2n 1 , n N2

But we know that cos cos

3 5cos cos cos .... cos 2n 1 0

2 2 2 2

2n 1 , n I2

Therefore, 2n 12

is the solution of equations cos = 0 for all numbers whose

cosine is 0.

17.5.3 To find a general solution of the equation tan = 0

It is given that tan 0

or sin 0cos

or sin 0

i.e. n , n I.

We have consider above the general solution of trigonometric equations, where the right handis zero. In the following, we take up cases where right hand side is non-zero.

17.5.4 To find the general solution of the equation sin = sin

It is given that sin sin

sin sin 0

or 2cos sin 02 2

Either cos 02

or sin 02

MATHEMATICS 129

Notes



2p 12 2

or q , p, q I2

2p 1 or 2q ....(1)

From (1), we get

nn 1 , n I as the geeneral solution of the equation sin sin

17.5.5 To find the general solution of the equation cos = cos

It is given that, cos cos

cos cos 0

2sin sin 02 2

Either, sin 02

or sin 02

p2

or q , p,q I2

2p or 2p ....(1)

From (1), we have

2n , n I as the general solution of the equation cos = cos

17.5.6 To find the general solution of the equation tan = tan

It is given that, tan tan

sin sin 0cos cos

sin cos sin cos 0

sin 0

n , n I

n n I

Similarly, for cosec = cosec , the general solution is

nn 1

and, for sec sec , the general solution is

2nand for cot cot

n is its general solution

MATHEMATICS

Notes


130


If 2 2sin sin , then

1 cos2 1 cos22 2

cos2 cos2

2 2n 2 , n I

n

Similarly, if 2 2cos cos , then

n , n I

Again, if 2 2tan tan , then

2 2

2 21 tan 1 tan1 tan 1 tan

cos2 cos2

2 2n 2

n , n I is the general solution.

Example 17.25 Find the general solution of the following equations :

(a) (i)1sin2

(ii)3sin

2

(b) (i)3cos

2(ii)

1cos2

(c) cot 3 (d) 24sin 1Solution :

(a) (i) 1sin sin2 6

nn 1 , n I6

(ii)3 4sin sin sin sin

2 3 3 3

n 4n 1 , n I3

(b) (i)3cos cos

2 6

MATHEMATICS 131

Notes



2n , n I6

(ii)1 2

cos cos cos cos2 3 3 3

22n , n I3

(c) cot 3

1 5tan tan tan tan

6 6 63

5n , n I6

(d) 24sin 12

2 21 1sin sin4 2 6

sin sin6

n , n I6

Example 17.26 Solve the following :

(a) 22cos 3sin 0 (b) cos4x cos2x

(c) cos3x sin2x (d) sin2x sin 4 x sin6x 0Solution :

(a) 22cos 3sin 0

22 1 sin 3sin 0

22sin 3sin 2 0

2sin 1 sin 2 0

1sin2

or sin 2

Since sin 2 is not possible.

7sin sin sin sin

6 6 6

n 7n 16

, n I

MATHEMATICS

Notes


132


(b) cos4x cos2xi.e., cos4x cos2x 0

2sin3xsinx 0

sin3x 0 or sinx 0

3x n or x n

nx3

or x n n I

(c) cos3x sin2x

cos3x cos 2x2

3x 2n 2x2 n I

Taking positive sign only, we have

3x 2n 2x2

5x 2n2

2nx5 10

Now taking negative sign, we have

3x 2n 2x2

x 2n2 n I

(d) sin2x sin 4 x sin6x 0

or sin6x sin2x sin4x 0

or 2sin4xcos2x sin 4 x 0

or sin4x 2cos2x 1 0

sin4x 0 or1 2cos2x cos2 3

4x n or22x 2n3

, n I

nx4

or x n3 n I

MATHEMATICS 133

Notes




1. Find the most general value of satisfying :

(i)3sin

2(ii) cosec 2

(ii)3sin

2(ii) 1

sin2


(i) 1cos2

(ii) 2sec

3

(iii) 3cos2

(iv) sec 2


(i) tan 1 (ii) tan 3 (iii) cot 1


(i) 1s in22

(ii) 1cos22

(iii) 1tan3

3

(iv) 3cos32

(v) 2 3sin4

(vi) 2 1sin 24

(vii) 24cos 1 (viii) 2 3cos 24

5. Find the general solution of the following :

(i) 22sin 3 cos 1 0 (ii) 24cos 4sin 1

(iii) cot tan 2 cosec

� sin A B sin A c o s B cosAsinB,

cos A B cosAcosB s inAs inB∓

tanA tan Btan A B1 t a n A t a n B

,tanA tan Btan A B

1 t a n A t a n B

LET US SUM UP

MATHEMATICS

Notes


134


cotAcotB 1cot A BcotB cotA

,cotAcotB 1cot A BcotB cot A

� 2sinAcosB sin A B sin A B

2cosAsinB sin A B sin A B

2cosAcosB cos A B cos A B

2sinAsinB cos A B cos A B

�C D C DsinC sin D 2sin cos

2 2C D C DsinC sin D 2cos sin

2 2

C D C DcosC cosD 2cos cos2 2

C D D CcosC cosD 2sin sin2 2

� 2 2sin A B sin A B sin A sin B

2 2cos A B cos A B cos A sin B

� 22 t a n Asin2A 2s inAcosA

1 tan A

�

22 2 2 2


� 22 t a n Atan2A

1 tan A

� 2 1 cos2Asin A2

, 2 1 cos2Acos A2

, 2 1 cos2Atan A1 cos2A

� 3sin3A 3sinA 4sin A , 3cos3A 4cos A 3cosA3

23tanA tan Atan3A

1 3tan A

� 3 3sinA sin 3Asin A4

, 3 3cosA cos3Acos A4

�A 1 cosAsin2 2

,A 1 cosAcos2 2

A 1 cosAtan2 1 cosA

MATHEMATICS 135

Notes



�5 1sin

10 4,

5 1cos5 4

� sin 0 n , n I

cos 0 2n 1 ,2 n I

tan 0 n , n I

� sin sin nn 1 , n I

� cos cos 2n , n I

� tan tan n , n I

� 2 2sin sin n , n I

� 2 2cos cos n , n I

� 2 2tan tan n , n I

NS


TERMINAL EXERCISE

1. Prove that 2 2

2 2cos B cos Atan A B tan A Bcos B sin A

2. Prove that cos 3sin 2cos3

3. If A B4

Prove that 1 tanA 1 tan B 2 and cot A 1 cosB 1 24. Prove each of the following :

(i)sin A B sin B C sin C A

0cosAcosB cosBcosC cosCcosA


MATHEMATICS

Notes


136


(ii)2 2

cos A cos A cos A cos A cos2A10 10 5 5

(iii) 2 4 9 1cos cos cos9 9 9 8

(iv) 13 17 43cos cos cos 045 45 45

(v) 1tan A cot A6 6 sin2A sin

3

(vi)sin s i n 2 tan

1 cos cos2(vii)

cos sin tan 2 sec2cos sin

(viii)2

21 sin tan1 sin 4 2

(ix) 2 2 2 3cos A cos A cos A

3 3 2

(x)sec8A 1 tan8Asec 4 A 1 tan2A

(xi)7 11 13 11cos cos cos cos

30 30 30 30 16

(xii)13 1sin sin

10 10 25. Find the general value of ' ' satisfying

(a)1

sin2 (b)

3sin2

(c)1

sin2 (d) cosec 2

6. Find the general value of ' ' satisfying

(a)1cos2

(b)2

sec3

(c)3cos

2(d) sec 2

7. Find the most general value of ' ' satisfying

(a) tan 1 (b) tan 1 (c) 1cot

38. Find the general value of ' ' satisfying

(a) 2 1sin2

(b) 24cos 1 (c) 2 22cot cosec

MATHEMATICS 137

Notes



9. Solve the following for :

(a) cosp cos q (b) sin9 sin

(c) tan5 cot

10. Solve the following for :

(a) sinm sin n 0 (b) tan m cotn 0

(c) cos cos2 cos3 0 (d) sin sin2 sin3 sin4 0

MATHEMATICS

Notes


138


ANSWERS


1. (a) (i)3 12 2

(ii)3

2(c)

21221

2. (a)3 12 2


1. (i)2 2cos A sin A

2(ii)

14

2. (c)3 12 2

CHECK YOUR PROGRESS 17.31. (a) sin5 sin ; (b) cos2 cos6

(c) cos cos3 6

(d) sin sin2 6

2. (a) 2sin5 cos (b) 2cos5 s in2(c) 2sin3 sin (d) 2cos6 cos


2. (a)2425

(b)120169

(c)20164225

3. (a)161289

(b)7

25(c)

119169

4. (a)247

(b) 2 22ab

b a5. 1


2. (a)2227

(b)2 3

3

3pq 4p

q3. (a)

2327

(b)3 2

34c 3cd

d


2. (a)3 12 2

,4 2 6

2 2

MATHEMATICS 139

Notes



3. (a) 2 22

(b) 2 22

(c) 2 1


1. (i)nn 1 , n I

3(ii)

nn 1 , n I4

(iii)n 4n 1 , n I

3(iv)

n 5n 1 , n I4

2. (i)22n , n I3

(ii)52n , n I6

(iii) 2n , n I6

(iv)32n , n I4

3. (i)3n , n I4

(ii) n , n I3

(iii) n , n I4

4. (i)nn 1 , n I

2 12(ii) n , n I

6

(iii)n , n I3 18

(iv)2n 5 , n I

3 18

(v) n , n I3

(vi)n , n I2 12

(vii) n , n I3

(viii)n , n I2 12

5. (i)52n , n I6

(ii)nn 1 , n I

6

(iii) 2n , n I3

TERMINAL EXERCISE

5. (a)nn 1 , n I

4(b)

nn 1 , n I3

(c)n 5n 1 , n I

4(d)

nn 1 , n I4

6. (a) 2n , n I3

(b) 2n , n I6

MATHEMATICS

Notes


140


(c)52n , n I6

(d)22n , n I3

7. (a) n , n I4

(b) 3n , n I4

(c) 2n , n I3

8. (a) n , n I4

(b) n , n I3

(c) n , n I4

9. (a) 2n ,n Ip q∓

(b)n or 2n 1 , n I4 10

(c) 2n 1 , n I12

10. (a)2k 1m n

or2k , k I

m n(b)

2k 1 ,k I2 m n

(c) 2n 14

or22n , n I3

(d)2n or n , n I

5 2 or 2n 1 ,n I

MATHEMATICS 141

Notes


Inverse Trigonometric Functions

18

INVERSE TRIGONOMETRIC FUNCTIONS

In the previous lesson, you have studied the definition of a function and different kinds of functions.We have defined inverse function.Let us briefly recall :Let f be a one-one onto function from A to B.Let y be an arbitary element of B. Then, f beingonto, an element x A such that f x y .Also, f being one-one, then x must be unique. Thusfor each y B , a unique element x A suchthat f x y . So we may define a function,denoted by 1f as 1f : B A

1f y x f x y

The above function 1f is called the inverse of f. A function is invertiable if and only if f isone-one onto.In this case the domain of 1f is the range of f and the range of 1f is the domain f.Let us take another example.We define a function : f : Car Registration No.If we write, g : Registration No. Car, we see that the domain of f is range of g and therange of f is domain of g.So, we say g is an inverse function of f, i.e., g = f 1.In this lesson, we will learn more about inverse trigonometric function, its domain and range, andsimplify expressions involving inverse trigonometric functions.

OBJECTIVESAfter studying this lesson, you will be able to :� define inverse trigonometric functions;

MATHEMATICS

Notes


142

Inverse Trigonometric Functions� state the condition for the inverse of trigonometric functions to exist;� define the principal value of inverse trigonometric functions;� find domain and range of inverse trigonometric functions;� state the properties of inverse trigonometric functions; and� simplify expressions involving inverse trigonometric functions.

EXPECTED BACKGROUND KNOWLEDGE� Knowledge of function and their types, domain and range of a function� Formulae for trigonometric functions of sum, difference, multiple and sub-multiples of

angles.

18.1 IS INVERSE OF EVERY FUNCTION POSSIBLE ?

Take two ordered pairs of a function 1x , y and 2x , y

If we invert them, we will get 1y , x and 2y , x

This is not a function because the first member of the two ordered pairs is the same.Now let us take another function :

sin ,12 ,

1sin ,4 2 and

3sin ,

3 2

Writing the inverse, we have

1,sin2 ,

1 ,sin42 and

3,sin

2 3

which is a function.Let us consider some examples from daily life.

f : Student Score in MathematicsDo you think f 1 will exist ?It may or may not be because the moment two students have the same score, f 1will cease to bea function. Because the first element in two or more ordered pairs will be the same. So weconclude that

every function is not invertible.

Example 18.1 If f : R R defined by 3(x) x 4f . What will be f 1 ?

Solution : In this case f is one-to-one and onto both.

f is invertible.

Let 3 4y x3 34 4y x x y

So f 1, inverse function of f i.e., f 1 3 4y y

MATHEMATICS 143

Notes


Inverse Trigonometric FunctionsThe functions that are one-to-one and onto will be invertible.

Let us extend this to trigonometry :Take y = sin x. Here domain is the set of all real numbers. Range is the set of all real numberslying between 1 and 1, including 1 and 1 i.e. 1 1y .

We know that there is a unique value of y for each given number x.In inverse process we wish to know a number corresponding to a particular value of the sine.

Suppose1sin2

y x

5 13sin sin sin sin ....6 6 6

x

x may have the values as 6

,56

,13 ....

6Thus there are infinite number of values of x.y = sin x can be represented as

1,6 2 ,

5 1 ,....,6 2

The inverse relation will be

1 ,,2 6

1 5 ,....,2 6

It is evident that it is not a function as first element of all the ordered pairs is 12

, which contradicts

the definition of a function.

Consider y = sin x, where x R (domain) and y [ 1, 1] or 1 y 1 which is called range.This is many-to-one and onto function, therefore it is not invertible.Can y = sin x be made invertible and how? Yes, if we restrict its domain in such a way that itbecomes one-to-one and onto taking x as

(i) x2 2

, y [ 1, 1] or

(ii)3 5x2 2

y [ 1, 1] or

(iii)5 3x2 2

y [ 1, 1] etc.

Now consider the inverse function 1siny x .

We know the domain and range of the function. We interchange domain and range for theinverse of the function. Therefore,

MATHEMATICS

Notes


144


(i)2 2

y x [ 1, 1] or

(ii)3 52 2

y x [ 1, 1] or

(iii)5 32 2

y x [ 1, 1] etc.

Here we take the least numerical value among all the values of the real number whose sine is x

which is called the principle value of 1sin x .

For this the only case is 2 2

y . Therefore, for principal value of 1siny x , the domain

is [ 1, 1] i.e. x [ 1, 1] and range is 2 2

y .

Similarly, we can discuss the other inverse trigonometric functions.Function Domain Range

(Principal value)

1. 1siny x [ 1, 1] ,2 2

2. 1cosy x [ 1, 1] [0, ]

3. 1tany x R ,2 2

4. 1coty x R [0, ]

5. 1secy x x 1 or x 1 0, ,2 2

6. 1cosecy x x 1 or x 1 , 0 0,2 2

18.2 GRAPH OF INVERSE TRIGONOMETRIC FUNCTIONS

18.2 Gr

1siny x 1cosy x

MATHEMATICS 145

Notes



1tany x 1coty x

1secy x 1cosecy xFig. 18.2

Example 18.2 Find the principal value of each of the following :

(i) 1 1sin

2(ii) 1 1cos

2(iii) 1 1

tan3

Solution : (i) Let 1 1sin

2

or 1sin sin42

or4

(ii) Let 1 1cos2

1 2cos cos cos2 3 3

or 23

(iii) Let 1 1tan

3or 1 tan

3 or tan tan

6

6

MATHEMATICS

Notes


146


Example 18.3 Find the principal value of each of the following :

(a) (i) 1 1cos

2(ii) 1tan ( 1)

(b) Find the value of the following using the principal value :

1 3sec cos2

Solution : (a) (i) Let1 1

cos2 , then

1 cos2

or cos cos4

4(ii) Let 1tan ( 1) , then

1 tan or tan tan4

4

(b) Let 1 3cos2

, then

3 cos2

or cos cos6

6

1 23sec sec seccos 6 32

Example 18.4 Simplify the following :

(i) 1cos sin x (ii) 1cot cosec x

Solution : (i) Let 1sin x

x sin

1 2 2cos sin x cos 1 sin 1 x

(ii) Let 1cosec x

x cosec

MATHEMATICS 147

Notes



Also 2cot cosec 1

2x 1


1. Find the principal value of each of the following :

(a) 1 3cos2

(b) 1cosec 2 (c) 1 3sin2

(d) 1tan 3 (e) 1cot 1

2. Evaluate each of the following :

(a) 1 1cos cos3 (b) 1cosec cosec

4(c)

1 2cos cosec3

(d) 1tan sec 2 (e) 1cosec cot 3

3. Simplify each of the following expressions :

(a) 1sec tan x (b) 1 xtan cosec2

(c) 1 2cot cosec x

(d) 1 2cos cot x (e) 1tan sin 1 x

18.3 PROPERTIES OF INVERSE TRIGONOMETRIC FUNCTIONS

Property 1 1sin sin ,2 2

Solution : Let sin x1sin x1sin sin

Also 1sin sin x x

Similarly, we can prove that

(i) 1cos cos , 0

(ii) 1tan tan ,2 2

Property 2 (i) 1 1 1cosec x sinx (ii) 1 1 1cot x tan

x

MATHEMATICS

Notes


148


(iii) 1 1 1sec x cosx

Solution : (i) Let 1cosec xx cosec1 sinx

1 1sinx

1 1 1cosec x sinx

(iii) 1sec xx sec1 cosx

or 1 1cosx

1 1 1sec x cosx

Property 3 (i) 1 1sin x sin x (ii) 1 1tan x tan x

(iii) 1 1cos x cos x

Solution : (i) Let 1sin x

x sin or x sin sin1sin x or 1sin x

or 1 1sin x sin x

(ii) Let 1tan x

x tan or x tan tan1tan x or 1 1tan x tan x

(iii) Let 1cos x

x cos or x cos cos1cos x

1 1cos x cos x

Property 4. (i) 1 1sin x cos x2

(ii) 1 1tan x cot x2

(iii) 1 1cosec x sec x2

(ii) Let 1cot xx cot1 tanx

1 1tanx

1 1 1cot x tanx

MATHEMATICS 149

Notes



Soluton : (i) 1 1sin x cos x2

Let 1sin x x sin cos2

or 1cos x2

1cos x2

or 1 1sin x cos x2

(ii) Let 1cot x x cot tan2

1tan x2

or 1tan x2

or 1 1cot x tan x2

(iii) Let 1cosec x

x cosec sec2

1sec x2

or 1sec x2

1 1cosec x sec x2

Property 5 (i) 1 1 1 x ytan x tan y tan1 xy

(ii) 1 1 1 x ytan x tan y tan1 xy

Solution : (i) Let 1tan x , 1tan y x tan , y tan

We have to prove that 1 1 1 x ytan x tan y tan1 xy

By substituting that above values on L.H.S. and R.H.S., we have

L.H.S. = and R.H.S. 1 tan tantan1 tan tan

1tan tan L.H.S. The result holds.

Simiarly (ii) can be proved.

MATHEMATICS

Notes


150


Property 6 2

1 1 1 12 2 2

2x 1 x 2x2tan x sin cos tan1 x 1 x 1 x

(i) (ii) (iii) (iv)Let x tanSubstituting in (i), (ii), (iii), and (iv) we get

1 12tan x 2 tan tan 2 ......(i)

1 12 2

2x 2 tansin sin

1 x 1 tan

12

2 tansin

sec1sin 2sin cos1sin sin2 2 .....(ii)

2 21 1

2 21 x 1 tancos cos1 x 1 tan

2 21

2 2cos sincoscos sin

1 2 2cos cos sin1cos cos2 2 ......(iii)

1 12 2

2x 2tantan tan

1 x 1 tan1tan tan 2 2 .....(iv)

From (i), (ii), (iii) and (iv), we get2

1 1 1 12 2 2


Property 7

(i)1 1 2 1

2

xsin x cos 1 x tan1 x

12

1sec1 x

21 1 x

cotx

1 1cosecx

MATHEMATICS 151

Notes



(ii)2

1 1 2 1 1 xcos x sin 1 x tan

x

12

1cosec1 x

12

xcot1 x

1 1secx

Proof : Let 1sin x sin x

(i) 2cos 1 x ,2

xtan1 x

,2

1sec1 x

,21 xcot

x and

1cosecx

1 1 2 12


12

1sec1 x

21 1 x

cotx

1 1cosecx

(ii) Let 1cos x x cos

2sin 1 x ,21 xtan

x,

1secx

, 2

xcot

1 x

and2

1cosec

1 x

1 1 2cos x sin 1 x

21 1 x

tanx

12

1cosec1 x

1 1secx

MATHEMATICS

Notes


152


Property 8

(i) 1 1 1 2 2sin x sin y sin x 1 y y 1 x

(ii) 1 1 1 2 2cos x cos y cos xy 1 x 1 y

(iii) 1 1 1 2 2sin x sin y sin x 1 y y 1 x

(iv) 1 1 1 2 2cos x cos y cos xy 1 x 1 y

Proof (i) : Let x sin , y sin , then

L.H.S. =

R.H.S. 1sin sin cos cos sin

1sin sin

L.H.S. = R.H.S.

(ii) Let x cos and y cos

L.H.S. =

R.H.S. 1cos cos cos sin sin

1cos cos

L.H.S. = R.H.S.

(iii) Let x sin , y sin

L.H.S. =

R.H.S. = 1 2 2sin x 1 y y 1 x

1 2 2sin sin 1 sin sin 1 sin

1sin sin cos cos sin1sin sin

L.H.S. = R.H.S.

(iv) Let x cos , y cos L.H.S =

R.H.S. = 1cos cos cos sin sin1cos cos

L.H.S. = R.H.S.

MATHEMATICS 153

Notes



Example 18.5 Evaluate :

1 13 5cos sin sin5 13

Soluton : Let 1 3sin5

and 1 5sin13 , then

3sin5

and 5sin

134cos5

and 12cos13

The given expression becomes cos

cos cos sin sin

4 12 3 5 335 13 5 13 65


1 1 11 1 2tan tan tan7 13 9

Solution : Applying the formula :

1 1 1 x ytan x tan y tan1 xy , we have

1 1 1 1 1

1 11 1 20 27 13tan tan tan tan tan1 17 13 90 91

7 13


1 1 14 12 33cos cos cos5 13 65

Applying the property

1 1 1 2 2cos x cos y cos xy 1 x 1 y , we have

1 1 14 12 4 12 16 144cos cos cos 1 15 13 5 13 25 169

1 33cos65

MATHEMATICS

Notes


154



1 11 1 xtan x cos2 1 x

Solution : Let x tan then

L.H.S. = and R.H.S. = 2

1 12

1 1 tan 1cos cos cos22 21 tan

1 22

L.H.S. = R.H.S. Example 18.9 Solve the equation

1 11 x 1tan tan x ,x 01 x 2

Solution : Let x tan , then

1 11 tan 1tan tan tan1 tan 2

1 1tan tan4 2

14 2

24 3 6

1x tan

6 3

Example 18.10 Show that

2 21 1 2

2 2

1 x 1 x 1tan cos x4 21 x 1 x

Solution : Let 2x cos2 , then

1 22 cos x

1 21 cos x2

Substituting 2x c o s 2 in L.H.S. of the given equation, we have

2 21 1

2 2

1 cos2 1 cos21 x 1 xtan tan1 cos2 1 cos21 x 1 x

MATHEMATICS 155

Notes



1 2cos 2sintan

2cos 2sin

1 cos sintancos sin

1 1 tantan1 tan

1tan tan4

4

1 21 cos x4 2



(a) 1 1sin sin3 2 (b) 1 1cot tan cot

(c)2

1 12 2

1 2x 1 ytan sin cos2 1 x 1 y

(d) 1 1tan 2tan5 (e) 1 1tan 2tan

5 4

2. If 1 1cos x cos y , prove that

2 2 2x 2xycos y sin

3. If 1 1 1cos x cos y cos z , prove that2 2 2x y z 2xyz 1

4. Prove each of the following :

(a)1 11 2sin sin

25 5 (b) 1 1 14 5 16sin sin sin5 13 65 2

(c) 1 1 14 3 27cos tan tan5 5 11

(d) 1 1 11 1 1tan tan tan2 5 8 4

MATHEMATICS

Notes


156


LET US SUM UP

� Inverse of a trigonometric function exists if we restrict the domain of it.

(i) 1sin x y if s iny x where 1 x 1, y2 2

(ii) 1cos x y if cos y = x where 1 x 1, 0 y

(iii) 1tan x y if tan y = x where x R, y2 2

(iv) 1cot x y if cot y = x where x R,0 y

(v) 1sec x y if sec y = x where x 1, 0 y2

or x 1, y2

(vi) 1cosec x y if cosec y = x where x 1, 0 y2

or x 1, y 02

� Graphs of inverse trigonometric functions can be represented in the given intervals byinterchanging the axes as in case of y = sin x, etc.

� Properties :

(i) 1sin sin , 1tan tan , 1tan tan and 1sin sin

(ii) 1 1 1cosec x sinx

, 1 1 1cot x tanx

, 1 1 1sec x cosx

(iii) 1 1sin x sin x , 1 1tan x tan x , 1 1cos x cos x

(iv) 1 1sin x cos x2

, 1 1tan x cot x2

, 1 1cosec x sec x2

(v) 1 1 1 x ytan x tan y tan1 xy

, 1 1 1 x ytan x tan y tan1 xy

(vi)2

1 1 1 12 2 2


(vii) 1 1 2 12


21 1 12

1 11 xsec cot cosecx1 x x

(viii) 1 1 1 2 2sin x sin y sin x 1 y y 1 x

(ix) 1 1 1 2 2cos x cos y cos xy y 1 x 1 y∓

LET US SUM UP

MATHEMATICS 157

Notes




TERMINAL EXERCISE


(a) 1 1 13 8 77sin sin sin5 17 85 (b) 1 1 11 1 1 3tan tan cos

4 9 2 5

(c) 1 1 14 3 27cos tan tan5 5 11


(a) 1 1 11 1 232tan tan tan2 5 11

(b) 1 1 11 1tan 2tan tan 22 3

(c) 1 1 11 1 1tan tan tan8 5 3

3. (a) Prove that 1 1 22sin x sin 2x 1 x

(b) Prove that 1 1 22cos x cos 2x 1

(c) Prove that 1 1 11 x 1 xcos x 2sin 2cos2 2


(a) 1 cosx xtan1 sin x 4 2

(b) 1 cosx sin xtan xcosx sin x 4

(c) 1 1 1ab 1 bc 1 ca 1cot cot cot 0a b b c c a

5. Solve each of the following :

(a) 1 1tan 2x tan 3x4

(b) 1 12tan cosx tan 2cosecx

(c) 1 1 1cos x sin x2 6 (d) 1 1cot x cot x 2 , x 0

12


MATHEMATICS

Notes


158


ANSWERS


1. (a)6

(b)4

(c)3

(d)3

(e)4

2. (a)13

(b)4

(c)12

(d) 1 (e) 2

3. (a) 21 x (b) 2

2

x 4(c) 4x 1 (d)

2

4

x

x 1(e)

1 xx


1. (a) 1 (b) 0 (c)x y

1 xy (d)5

12(e)

717

TERMINAL EXERCISE

5. (a)16

(b)4

(c) 1 (d) 3

158/1

Hyperbolic Functions

Objectives

If we take x = a cos θ, y = asinθ (θ∈ R) Then x2 + y2 = a2, In other words, for any real value

of θ the point (a cos θ, a sinθ) lies on the circle x2 + y2 = a2. For this reason, the trigonometric

function we are also known as circular functions.

,2 2

e e e ex a y b

θ −θ θ −θ + += = (θ∈ R) then we get that

2 2

2 21

x y

a b− = . This is the

equation of "hyperbpla" (Hyperbolas will be discussed in the second year intermediate course). This

mean that points on the hyperbola are in the 2 2

2 21

x y

a b− = form

, ( R)2 2

e e e ea b

θ −θ θ −θ + − θ ∈

Keeping this fact in view, the "Hyperbolic functions" are introduced. The number e is defined as

0

1 1 11 ..... R

1 2 !n

e xn

∞

== + + + ∞ = ∀ ∈∑

For any it is known that

2

0

1 .....1 2 !

nx

n

x x xe

n

∞

== + + + ∞ = ∑

e is also give by 1

lt 1x

ne

x→∞

= +

the approximate value of e is given by e~ 2.7118287.

158/2

18.4 Definition

18.4.1Definition : 1) The functionf : R→R defined by ( )2

x xe ef x

−−= for all Rx∈ is called

the "hyperbolic sine" function. It is denoted by sin hx.

Thus sinh2

−−=

x xe ex for all Rx∈

ii) cosh2

−+=x xe e

x for all Rx∈

iii)sinh

tanhcosh

−

−−= =+

x x

x x

x e ex

x e e for all Rx∈

iv)cosh

cot hsinh

−

−+= =−

x x

x x

x e ex

x e e for all Rx∈

v)1 2

sechcosh −= =

+x xxx e e

for all Rx∈

vi)1 2

cosechsinh −= =

−x xxx e e

for all Rx∈

18.4.2 Identities

i) cosh2x − sinh2x = 1 for all Rx∈

ii) 1 − tanh2x = sech2x for all Rx∈

iii) coth2x − 1 = cosech2x for all Rx∈ − {0}

158/3

18.5 Inverse Hyperbolic function

18.5.1 Definition

The function f : R→R defined by f (x) = sinh x nfor all Rx∈ , is a bijection. Thus the inverse

of this function exists and it is denoted by sinh−1 . Thus, if x, y are real numbers then

sinh−1x = y ⇔ sinhy = x

⇒ cosh−1x = y ⇔ coshy = x

tanh−1x = y ⇔ tanhy = x

coth−1x = y ⇔ cothy = x

sech−1x = y ⇔ sechy = x

cosech−1x = y ⇔ cosechy = x we define.

18.6 Properties of hyperbolic function

18.6.1 for x, y ∈ R

i) sinh(x + y) = sinhx coshy + coshx sin hy

ii) sinh (x− y) = sinhx coshy − coshx coshy

iii) cosh(x + y) = coshx coshy + sinhx sinhy

iv) cos(x −y) = cohshx coshy − sinhx sinhy

18.6.2 for any x ∈ R

i) sinh2x = 2sinhx coshx = 2

2 tanh

1 tanh−x

x

ii) cosh 2x = cosh2x + sinh2x

= 2 cosh2x − 1

= 1 + 2 sinh2x

=2

2

1 tanh

1 tanh

+−

x

x

158/4

18.6.3 for any x, y ∈ R

i) tanh (x + y) = tanh tanh

1 tanh tanh

++

x y

x y

ii) tanh(x − y) = tanh tanh

1 tanh tanh

−−

x y

x y

iii) coth (x + y) = coth coth 1

coth cot h

++

x y

y x

iv) coth (x − y) = coth coth 1

coth cot h

−−

x y

y x

18.6.4 for any x ∈ R

i) tanh 2x = 2

2 tanh

1 tanh+x

x

ii) coth 2x = 2coth 1

2coth

+x

x if x ≠ 0

Example 1 : Prove that, for any x ∈ R sinh(3x) = 3sinhx + 4 sinh3x

Sol: sinh3x = sinh (2x + x)

= sinh2x . coshx + cosh 2x sin hx

= 2 sinhx (1 + sinh2x) + (1 + 2sinh2x) sinhx

= 3sinhx + 4sinh3x

Example 2: If 5cosh

2=x find the value of i) cosh(2x) ii) sinh(2x)

Sol : i) cosh(2x) = 2cosh2x − 1

= 25 23

2. 14 2

− =

ii) we know that cosh2(2x) − sinh2(2x) = 1

158/5

Therefore2

2 2 23sinh (2 ) cosh (2 ) 1 1

2 = − = −

x x

25.24

4=

5 21sinh(2 )

2= ±x

Check your progress 1

1. If3

sinh4

=x find the value of cosh(2x), sinh(2x).

2. Prove that tanh tanhtanh( )

1 tanh tanh

−− =−

x yx y

x y.

3. Prove that cosh4(x) − sinh4x = cosh(2x).

Letus sum up

1. i) sinh ,2

−−=

x xe ex

ii) cosh2

−+=

x xe ex

2. i) sinh

tanh ;cosh

= xx

x

ii) cosh 1

cothsinh tanh

= =xx

x x

iii)1

sec ;cosh

=hxx

1cosec

sinh=hx

x

3. i) cosh2x − sinh2x = 1

ii) 1 − tanh2x = sech2x

iii) coth2x− 1 = cosech2x.

158/6

4. i) sinh 2x = 2sinhx coshx,

ii) cosh 2x = cosh2x + sinh2x

5. i) sinh3x = 3sinhx+ 4sinh3x

ii) cosh 3x = 4cosh3x − 3coshx

iii)3

3

3tanh tanhtanh 3

1 tanh

+=+

x xx

x

Supportive Websites

� http://www.wikipedia.org

� http://mathworld.wolfram.com

Answers


1.17 15

,8 8

MATHEMATICS 159

Notes


Relations Between Sides and Angles of a Triangle

19

RELATIONS BETWEEN SIDES ANDANGLES OF A TRIANGLE

In an earlier lesson, we have learnt about trigonometric functions of real numbers, relationsbetween them, drawn the graphs of trigonometric functions, studied the characteristics fromtheir graphs, studied about trigonometric functions of sum and difference of real numbers, anddeduced trigonometric functions of multiple and sub-multiples of real numbers. We also studiedabout inverse trigonometric functions, some of their properties and solved problems based ontheir concepts.In this lesson, we shall try to establish some results which will give the relationship between sidesand angles of a triangle and will help in finding unknown parts of a triangle.

OBJECTIVES

After studying this lesson, you will be able to :� derive sine formula, cosine formula and projection formula� apply these formulae to solve problems.

EXPECTED BACKGROUND KNOWLEDGE� Trigonometric and inverse trigonometric functions.� Formulae for sum and difference of trigonometric functions of real numbers.� Trigonometric functions of multiples and sub-multiples of real numbers.

19.1 SINE FORMULAIn a ABC, the angles corresponding to the vertices A, B, and C are denoted by A, B, and Cand the sides opposite to these vertices are denoted by a, b and c respectively. These anglesand sides are called six elements of the triangle.

MATHEMATICS

Notes


160

Relations Between Sides and Angles of a TriangleResult 1 : Prove that in any triangle, the lengths of the sides are proportional to the sines of theangles opposite to the sides,

i.e.a b c

sinA sinB sinC

Proof : In ABC, in Fig. 19.1 [(i), (ii) and (iii)], BC = a, CA = b and AB = c and C is acuteangle in (i), right angle in (ii) and obtuse angle in (iii).

(i) (ii) (iii)Fig. 19.1

Draw AD prependicular to BC (or BC produced, if need be)

In ABC,AD sinBAB

orAD sinBc

AD = c sin B .....(i)

In ADC, AD sinCAC

in Fig 19.1 (i)

or,AD sinCb

AD = b sin C ....(ii)

If Fig. 19.1 (ii), or AD 1 sin sinCAC 2AD = b sin C

and in Fig. 19.1 (iii), AD sin C sinCAC

orAD sinCb

or AD = b sin C

Thus, in all three figures, AD = b sin C

From (i) and (ii), we get

c sin B = b sin C

b csinB sinC

....(iii)

MATHEMATICS 161

Notes



Similarly, by drawing prependicular from C on AB, we can prove thata b

sinA sinB.....(iv)

From (iii) and (iv), we geta b c

sinA sinB sinC.....(A)

(A) is called the sine-formula

Note : (A) is sometimes written assinA sinB sinC

a b c....(A')

The relations (A) and (A') help us in finding unknown angles and sides, when some others aregiven.

Let us take some examples :

Example 19.1 Prove that B C Aacos b c sin

2 2, using sine-formula.

Solution : R.H.S.Ab c sin2

We know that, a b c k

sinA sinB sinC (say)

a = k sin A, b = k sin B, c = k sin C

R.H.S.Ak sinB sinC sin2

B C B C Ak 2 sin cos sin2 2 2

NowB C A90

2 2A B C∵

B C Asin cos2 2

R.H.S.A B C A2 k cos cos sin2 2 2

B Ck sinA cos2

= B Ca cos L.H.S

2

MATHEMATICS

Notes


162


Example 19.2 Using sine formula, prove that

2 Aa cosC cosB 2 b c cos2

Solution : We havea b c k

sinA sinB sinC(say)

a = k sin A, b = k sin B, c = k sin C

R.H.S 2 A2k sinB sinC cos2

2B C B C A2k 2cos sin cos2 2 2

2A B C A B C A4ksin sin cos 2asin cos2 2 2 2 2B C B C2asin sin

2 2

a cosC cosB= L.H.S.

Example 19.3 In any triangle ABC, show that

asinA bsinB csin A B

Solution : We havea b c k

sinA sinB sinC (say)

L.H.S. = ksinA sinA ksinB sinB

2 2k sin A sin B

ksin A B sin A B

A B C sin A B sinC

L.H.S. ksinC sin A B

csin A B R.H.S.

Example 19.4 In any triangle, show that

2 2a b cosC ccosB b c

Solution : We have, a b c ksinA sinB sinC

(say)

MATHEMATICS 163

Notes



L.H.S. = k sinA k sinBcosC k sinCcosB

2k sinA sin B C

2k sin B C sin B C sin A sin B C∵

2 2 2k sin B sin C2 2 2 2k sin B k sin C2 2b c R.H.S


1. Using sine-formula, show that each of the following hold :

(i)

A Btan a b2A B a btan

2

(ii) bcosB ccosC acos B C

(iii)B C Aasin cosb c

2 2(iv)

b c B C B Ctan cotb c 2 2

(v) acosA bcosB ccosC 2asinBsinC

2. In any triangle if a b

cosA cosB, prove that the tiangle is isosceles.

19.2 COSINE FORMULAResult 2 : In any triangle, prove that

(i)2 2 2b c acosA

2bc (ii)

2 2 2c a bcosB2ac

(iii)2 2 2a b ccosC

2abProof :

(i) (ii) (iii)Fig. 19.2

MATHEMATICS

Notes


164


Three cases arise :

(i) When C is acute (ii) When C is a right angle

(iii) When C is obtuse

Let us consider these one by one :

Case (i) When C is acute

AD sinCAC

AD b sin C

Also BD BC DC a bcosCDC cosCb

∵

From Fig. 19.2 (i)

2 2 2c bsinC a bcosC

2 2 2 2 2b sin C a b cos C 2abcosC

2 2a b 2abcosC

2 2 2a b ccosC2ab

Case (ii) When C = 90°

2 2 2 2 2c AD BD b aAs C = 90° cos C = 0

2 2 2c b a 2ab cosC2 2 2b a ccosC

2ab

Case (iii) When C is obtuse

AD sin sinC180 CACAD bsinC

Also, BD BC CD a bcos 180 Ca bcosC

2 22c bsinC a b cosC

2 2a b 2ab cosC

2 2 2a b ccosC2ab

MATHEMATICS 165

Notes



In all the three cases, 2 2 2a b ccosC

2abSimilarly, it can be proved that

2 2 2c a bcosB2ac

and2 2 2b c acosA

2bcLet us take some examples to show its application.

Example 19.5 In any triangle ABC, show that2 2 2cosA cosB cosC a b c

a b c 2abc

Solution : We know that2 2 2b c acosA

2bc,

2 2 2c a bcosB2ac

,2 2 2a b ccosC

2ab

L.H.S. 2 2 2 2 2 2 2 2 2b c a c a b a b c

2abc 2abc 2abc

2 2 2 2 2 2 2 2 21b c a c a b a b c2abc

2 2 2a b c R.H.S.2abc

Example 19.6 If A 60 , show that in ABC

(a b c)(b c a) 3bc

Solution : 2 2 2b c acosA

2bc.....(i)

A = 60° 1cosA cos602

(i) becomes 2 2 21 b c a

2 2bc

2 2 2b c a bc

or 2 2 2b c 2bc a 3bc

or 22 a 3bcb c

or 3bcb c a b c a .

MATHEMATICS

Notes


166


Example 19.7 If the sides of a triangle are 3 cm, 5 cm and 7 cm find the greatest angle of a

triangle.

Solution : Here a = 3 cm, b = 5 cm, c = 7 cm

We know that in a triangle, the angle opposite to the largest side is greatest

C is the greatest angle.

2 2 2a b ccosC2ab

9 25 49 15 130 30 2

1cosC2

2C3

The greatest angle of the triangle is 23

or 120°.

Example 19.8 In ABC, if A 60 , prove that b c 1

c a a b.

Solution :2 2 2b c acosA

2bc

or2 2 21 b c acos60

2 2bc

2 2 2b c a bc

or 2 2 2b c a bc .....(i)

2 2b c ab b c acR.H.S.c a a b c a a b

2ab ac a bcc a a b

[Using (i)]

a ca b a ba c a b

a c a b 1a c a b

MATHEMATICS 167

Notes




1. In any triangle ABC, show that

(i)2 2 2 2 2 2

2 2 2b c c a a bsin2A s i n 2 B sin2C 0

a b c

(ii) 2 2 2 2 2 2 2 2 2tanB tanC tanAa b c b c a c a b

(iii)2 2 2k a b c

sin2A sin2B sin2C2 2abc

wherea b c k

sinA sinB sinC

(iv) 2 2 2 2 2 2cotA cot B cotC 0b c c a a b2. The sides of a triangle are a = 9 cm, b = 8 cm, c = 4 cm. Show that

6 cos C = 4 + 3 cos B. 19.3 PROJECTION FORMULAResult 3 : In ABC, if BC = a, CA = b and AB = c, then prove that

(i) a b cosC c c o s B (ii) b c c o s A a c o s C (iii) c a c o s B b c o s A

Proof :

(i) (ii) (iii)Fig. 19.3

As in previous result, three cases arise. We will discuss them one by one.

(i) When C is acute :

In ADB ,BD cosBc

BD ccosB

In ADC ,DC cosCb

DC = b cos C

a = BD + DC = c cos B + b cos C

a = c cos B + b cos c

MATHEMATICS

Notes


168


(ii) When C 90BCa BC AB cosB cAB

ccosB 0

ccosB b cos90 cos90 0∵

ccosB b cosC

(iii) When C is obtuse

In ADB ,BD cosBc

BD = c cos B

In ADC ,CD cos cosCC

bCD bcosC

In Fig.19.3 (iii),BC BD CD

a ccosB bcosC

ccosB b cosC

Thus in all cases,a bcosC ccosBSimilarly, we can prove that

b ccosA acosC and c acosB bcosALet us take some examples, to show the application of these results.

Example 19.9 In any triangle ABC, show thatcosA cosB cosC a b cb c c a a b

Solution : L.H.S. bcosA ccosA ccosB acosB acosC b cosC

b cosA acosB ccosA a cosC ccosB b cosC

c b aa b c R.H.S.

Example 19.10 In any ABC , prove that

2 2 2 2cos2A cos2B 1 1

a b a b

Solution : L.H.S. 2 2

2 21 2sin A 1 2sin B

a b2 2

2 2 2 21 2sin A 1 2sin B

a a b b

2 22 2 2 2

1 1 1 12k 2ka b a b

sinA sinB ka b

= R. H. S.

MATHEMATICS 169

Notes



Example 19.11 In ABC , if acosA bcosB , where a b prove that ABC is a

right angled triangle.

Solution : acosA bcosB

2 2 2 2 2 2b c a a c ba b2bc 2ac

or 2 22 2 2 2 2 2a bb c a a c b

or 2 2 2 2 4 2 2 2 2 4a b a c a a b b c b

or 2 2 2 2 2 2 2c a b a b a b2 2 2c a b

ABC is a right triangle.

Example 19.12 If a = 2, b = 3, c = 4, find cos A, cos B and cos C.

Solution : 2 2 2b c acosA

2bc

9 16 4 21 72 3 4 24 82 2 2c a b 16 4 9 11cosB

2ac 2 4 2 16

and2 2 2a b c 4 9 16 3 1cosC

2ab 2 2 3 12 4


1. If a = 3, b = 4 and c = 5, find cos A, cos B and cos C.

2. The sides of a triangle are 7 cm, 4 3 cm and 13cm . Find the smallest angle of thetriangle.

3. If a : b : c = 7 : 8 : 9, prove thatcos A : cos B : cos C = 14 : 11 : 6.

4. If the sides of a triangle are 2x x 1 , 2x 1 and 2x 1. Show that the greatest angleof the triangle is 120°.

5. In a triangle, b cos A = a cos B, prove that the triangle is isosceles.

6. Deduce sine formula from the projection formula.

MATHEMATICS

Notes


170


It is possible to find out the unknown elements of a triangle, if the relevent elements are given byusing

Sine-formula :

(i)a b c

sinA sinB sinC

Cosine foumulae :

(ii)2 2 2b c acosA

2bc

2 2 2c a bcosB2ac

2 2 2a b ccosC2ab

Projection formulae :a bcosC ccosBb ccosA acosCc acosB bcosA


TERMINAL EXERCISE

In a triangle ABC, prove the following (1-10) :

1. a sin B C bsin C A csin A B 0

2. a c o s A bcosB ccosC 2asinBsinC

3.2 2 2 2 2 2

2 2 2b c c a a bsin2A sin2B sin2C 0

a b c

LET US SUM UP


MATHEMATICS 171

Notes



4.2 2

2 2c a 1 cosBcos C A

1 cosAcos B Cb c

5.c bcosA cosBb ccosA cosC

6.a bcosC s i n Cc bcosA sin A

7.A B Ca b c tan tan 2c cot2 2 2

8.A B a b Csin cos

2 c 2

9. (i) bcosB ccosC acos B C

(ii) acosA bcosB ccos A B

10.2 22 2 2B Bb c a cos c a sin

2 2

11. In a triangle, if b = 5, c = 6, A 1tan2 2

, then show thata 41 .

12. In any ABC , show that

cosA b acosCcosB a b cosC

MATHEMATICS

Notes


172


ANSWERS


1.4cosA5

3cosB5

cosC zero

2. The smallest angle of the triangle is 30°.

174/1

Heights and Distances

Objectives

We have studied about triangles in general and right angled triangles in particular in this chapter

we will combine the principles of both these topics to study some applications of trigonometry to

problems of daily life and some problems of scientific interest.

Suppose a boy is sitting on the balcony of this building located on one bank of a canal. He is

looking at a small pup standing on a step of the temple on the other bank of the canal. By Imagining a

right angled triangle to fit with the geometry of the situation and knowing the height at which the boy is

sitting. Can you determine the width of the canal.

To solve such problems we introduce two fundamental concepts - i) angle of elevation and ii)

angle of depression.

19.4 Angle of elevation and angle of depression

In this discussion that follows in this section, we discuss some elementary ideas like obesver's

line of right, the angle of elevation and the angle of depression and illustrate then diagrams. Later we

shall discuss some physical problem involving single plane which are based on these concepts and the

trigonometric ratios and obtain their solutions.

174/2

19.4.1 Definition (Angle of elevation)

The line of right is the line drawn from the eye of an observer to the object viewed by the

obeserver.

The anlge of elevation is defined as the anlge AOB formed by the horizontal OA with the line

of right OB, when the object being viewed is above the horizontal level.

Fig. 19.4

19.4.2 Definition (Angle of depresition)

The angle of depression of an object being viewed is defined as the angleCBOformed by the

horizontal CB with the line of sight BO when the object viewed is below the horizontal level.

Fig. 19.5

B

AO

horizontalobeserver's position

Line of sight

Angle of tertion

B

AO

horizontalobserving position

Line of sight

A of depB

C

174/3

Example 1

From a point on the level ground, the angle of elevation of the top of a hill is α. After moving

through a distance of 'd' meters towards the foot of the hill. The angle of elevation of its top is found

to be β. Find the height of the hill.

Sol: As shown in fig 19.5 Let 'B' be the foot of the hill AB. Let C be the initial point of observation and

D be the final point of observation on the horizontal. Let 'h' be the height of the hill and DB = x.

Fig. 19.5

By hypothesis CD = d ACB , ADB= α = β

From ∆ ACB, cot α = CB

AB

x d

h

+=

∴ x + d = h cot α ... (1)

From∆ ADB cotβ = DB

cotAB

xx h

h= ∴ = β ... (2)

From equation (1) and (2) we have d = h(cot α − cot β) ... (3)

cot cot

dh =

α − β

Height of the hill cot cot

d=α − β

A

BOα

xDd

β

174/4

Example 2

From the cliff of a mountain, 120 meters above the sea level, the angles of depression of two

boats of the same side of the mountain are 450 and 350 respectively. Find the distance between the

two boats. (Assume that the boats lie along a straight line from the foot of the hill)

Sol : Take MN as the mountain and consider a horizontal MX through the foot of the mountian M. Thehorizontal MX can be taken as the sea level. Draw NY parallel to the horizontal line such that it passes

through the observers eye. Denote the two boats by A and B.

Fig. 19.6

By hypothesis 0 0YNA 45 ; YNB 30= =

By from fig 11.4, YNA NAN=

From ∆ MBN, cot 300 = 0BM BM

= BM 120 cot 30MN 120

⇒ =

From∆ MAN , cot 450 = 0 0AM AM

= AM 120 cot 45MN 120

⇒ =

AB = BM − AM = 120 (cot 300 − cot 450)

= 120 ( 3 −1) mts.

Example 3

One end of the ladder is in contact with a wall and the other and is in count with the level ground

making an angle α . When the foot of the ladder is moved through a distance 'a' cms away from thewall, the end in contact with the walls slides through 'b' cms along the wall and the angle made by the

ladder with the level grounds is now β . Show that Tan2

a bα +β =

N

MBx

Ad

300450

450300

y

120 m

174/5

Sol: Let l be the length of the ladder, OA = x, OB = y. Let the intiati position of the ladder be AB and

the final position after its foot moved through a distance 'a' cms be A' B' . Then AB = A'B' = l.

Fig. 19.7

From fig 19.7 OB = OB − BB' = y − b

Also from ∆ OAB, cos cosx

x ll

α = ⇒ = α ...(1)

similarly OB = y = l sin α ...(2)

In ∆ OA'B' , OA' = x + a = l cos β ...(3)

and OB' = y − b = l sin β ...(4)

we now make out a relation between a, b, c, d using relation (1) - (4)

From (1) and (3); a = l (cos β − cos α)

From (2) and (4) ;b = l (sin α − sin β)

2sin sincos cos 2 2

Tansin sin 2

2 sin cos2 2

a

b

α −β α +β β− α α +β = = = α −β α + βα − β

Tan Tan2 2

aa b

b

α + β α + β = ⇒ =

B

A 'Oa

B'

x

αA

y−b

b

β

y

174/6


1. From the top of a light house 200 meters high on the sea - shore, the angle of depression of a ship

in the sea is 360. How far is the ship from the sea - shore ?

2. The angles of elevation of the top of a tower from two points at a distance of 4m and 9 m. From

the base of the tower and in the same straight line with it are complementary. Prove that the

height of the tower is 6m.

3. The angle of elevation of the top of a building from the foot of the tower is 300 and the angle of

elevation of the top of the tower from the foot of the building is 600. If the tower is 50 meters high,

find the hight of the buildings.

Letus sum up

1. The line of right is the line drawn from the eye of the observe to the object that is viewed by the

observer.

2. The angle of elevation of an object viewed, is the angle formed by the line of sight with the

horizontal, when the object that viewed is above the horizontal level. In this case, the observe

raises his/her head to look at the object.

3. The angle of depression of an object viewed is the angle formed by the line of sight with the

horizontal, when the object that viewed is below the horizontal level. In this case, the observer

lowers his/her hand to look at the object.

Supportive Websites

� http:// www.wikipedia.org

� http:// mathworld.wolfram.com.

Answers


(1) 275.26 meters

(3)50

3 meters

174/7

De Moivre's Theorem

Objectives

We learn that cisθ1 . cisθ2 = cis(θ1 + θ2) and hence (cisθ)2 = cis 2θ. In this chapter we extend

this result for any integer. The extension is called De Moivre's theorem for integral indices. We use it

in studing the nth roots of unity. We also obtain a verson of De Moivre's theorem for rational indices.

19.5 De Moivre's theorem for Integral Index and for rational Index

By using this theorem all the nth roots of a complex number Z ≠ 0 are determined. As a

particular case, all the nth roots of unity are determined and their geometrical representations are

obtained.

19.5.1 De Moivre's Theorem for Integral Index

for any real numberθ and any Integer n,

(cosθ + i sin θ)n = cos nθ + i sin nθ

19.5.2

i) It is customary to write cis θ for cos θ + i sin θ. Thus we may state the DeMoivre's theorm as

(cisθ)n ́ cis (nθ) if n ∈ Z.

ii) (cosθ + i sin θ)−n = cos (−n) θ + i sin (−n) θ = cos nθ − i sin nθ provided 'n' is an integer.

iii) (cosθ + i sin θ) (cos θ − i sinθ) = cos2 θ − i2 sin2 θ

= cos2 θ + sin2 θ = 1 (i2 = −1)

174/8

iv) (cos θ − i sin θ)n = 1

cos sin

n

i

θ + θ

= (cos θ + i sinθ)−n = cos nθ − i sin2 nθ

v) cis θ .cisϕ = cis (θ + ϕ) for any θ, ϕ ∈ R.

Example 1 :If m, n are integers and x = cos α + i sin α

Then prove thaty = cos β + i sin β then 1cos ( )m n

m nx y m nx y

+ = α + β

12 sin ( )m n

m nx y i m n

x y− = α + β

Sol : xm = (cos α + i sin α)m = cos mα + i sin mα

yn = (cos β + i sin β)n = cos nβ + i sin nβ

∴ xm yn= (cos mα + i sin mα) (cosnβ + i sin nβ)

= cos (mα + nβ) + i sin (mα + nβ) ...(1)

1 1cos ( ) sin ( )

cos ( ) sin ( )m nm n i m n

m n i m nx y= = α + β − α + β

α + β + α + β ...(2)

(1) + (2)1

2 cos ( )m nm n

x y m nx y

+ = α + β

(1) − (2) 1

2 sin ( )m nm nx y i m n

x y− = α + β

Example 2 :

If n is a positive integer, show that 2

2(1 ) (1 ) 2 cos4

nn n n

i i+ π + + − =

.

Sol : 1 11 2 2 cos sin

4 42 2i i i

π π + = + = +

1 11 2 2 cos sin

4 42 2i i i

π π − = − = −

2(1 ) ( 2) cos 2 cos sin4 4

n nn n n n

i i in n

π π π π + = + = + ...(1)

174/9

2(1 ) ( 2) cos 2 cos sin4 4

n nn n n n

i i in n

π π π π − = − = − ...(2)

By adding (1) & (2) we get

22 2(1 ) (1 ) 2 2cos 2 cos

4 4

n nn n n n

i i+π π + + − = =

Example 3:

If n is an integer then show that (l + cos θ + i sin θ)n + (1 + cos θ − i sin θ)n

12 cos cos2 2

n n n+ θ θ =

Sol : Take L.H.S 2(1 cos sin ) (1 cos sin ) 2cos 2 sin cos

2 2 2

nn ni i i

θ θ θ + θ + θ + + θ− θ = +

22cos 2 sin cos2 2 2

n

iθ θ θ + −

2 cos cos sin cos sin2 2 2 2 2

n nn n i i

θ θ θ θ θ = + + −

2 cos cos sin cos sin2 2 2 2 2

n n n n n ni i

θ θ θ θ θ = + + −

12 cos 2cos 2 cos cos2 2 2 2

n n n nn n+θ θ θ θ = = .

Check Your Progress 1

1) i) 3(1 3)i+ iv)

5 53 3

2 2 2 2i i

+ − − find the values

2) If cos α + cos β + cos γ = 0 = sin α + sin β + sin γ show that

i) cos 3α + cos 3β + cos 3γ = 3 cos (α + β + γ)

ii) sin 3α + sin 3β + sin γ = 3 sin (α + β + γ)

174/10

19.6 nth roots of Unity

Letn be a positive integer greater than one and 2 2

cos sinw in n

π π= + , then 1, w, w2 ...

wn−1 are all the distinct nth roots of unity.

i) The sum of the nth roots of unity is zero

1 + w + w2 + .... + wn−1 = 0

ii) product of the nth roots of unity is (−1)n−1

19.7 Cube roots of Unity

In this section we find all the cube roots of unity and all the cube roots of a positive real numbers.

19.7.1

1,1 3 1 3

,2 2

i i− + − − are the cube roots of the unity

w = cis 2

n

π .

Example 4 : Find the values of ( )143 i+ .

Sol : The modulus ampllitude form of 3 i+

= 2(cos 300 + i sin 300)

Hence( )1

1 14

4 43 2 cos sin6 6

i iπ π + = +

14

262 cis , 0,1, 2, 3

6

kk

π π + = =

14 12

2 cis , 0,1, 2, 324

kk

π + π = =

142 cis (12 1) , 0,1, 2, 3

24k k

π= + =

174/11

( )143 i+ is all values

1 1 1 14 4 4 413 25 37

2 cis , 2 cis , 2 cis , 2 cis24 4 24 4

π π π π

Example 5.

Find all the roots of the equationx11 − x7 + x4 − 1 = 0

Sol : x11 − x7 + x4 − 1 = x7 (x4 − 1) + 1(x4 − 1)

= (x4 − 1) (x7 + 1)

Therefore the roots of the given equations are precisely the 4th roots of unity and 7th roots of

−1. They are

2cis cis {0, 1, 2, 3}

4 2k k

kπ π= ∈ and

2cis cis (2 1) , {0, 1, 2, 3, 4, 5, 6}

7 7

kk k

π + π π= + ∈

The value of x are

cis 0 = 1, cis 2

π = i, cis π = −1 [� −1 = cis π]

3cis

2i

π = −

3 5 9 11 13cis , cis , cis , cis 1, cis , cis , cis

7 7 7 7 7 7

π π π π π ππ=−

i.e., 3 5 9 11 13

i, 1, cis , cis , cis , cis , cis , cis7 7 7 7 7 7

π π π π π π± ±

Example 6

If 1, w, w2 are the cube roots of unity, prove that x2 + 4x + 7 = 0 where x = w w2 − 2

Sol : x = w − w2 − 2

⇒ (x + 2) = w − w2

⇒ (x + 2)2 = w2 + w4 − 2 w3

⇒ x2 + 4x + 4 = w2 + w − 2 = −1 −2 = −3

⇒ x2 + 4x + 7 = 0

174/12

Check your Progress 2

1. Find all the value of (−16)1/4.

2. If 1, w, w2 are the cube roots of unity the prove that 1 1 1

2 1 2 1w w w+ =

+ + + .

3. Solve x4 −1 = 0 the equation.

4. Solve the equationx9 − x5 + x4 −1 = 0.

Letus sum up

1. If 'n' is an inteter, then (cisθ)n = cis (nθ) [De-Moiver's theorem for integral index]

2. If Z0 = r0 cis θ0 ≠ 0 then the nth roots of Z0are

10

0

2cis , 0, 1, 2, ....( 1)n

kkn

a r k nn

+ θ = = −

3. The nth roots of unity are cis 2k

n

πk = 0, 1, 2, 3, .... (n − 1)

4. Cube roots of unity are 21 3 1 1 31, ,

2 2

iw w

− + − −= =

Suportive Websites



Answers


(1) 8 (3) i


1) 2 cis (2k + 1) 4

πk = 0, 1, 2, 3

3) + 1, + i

4) + 1, + i, cis 3

, cis4 5

π π ± ±

174/13

Trigonometric Expansions

Objectives

In this section we discuss expansions for sin(nθ) cos(nθ) and tan (nθ) in terms of powers of sin

θ, cosθ, tanθ when 'n' is a positive integer. These expressions are consequence of De-Moiver's

theorem and have finitely many terms.

19.8 Expansion of trigonometric functions sin nθθθθθ, cos nθθθθθ, tan nθθθθθ, cot nθθθθθ

Here we are De Moiver's theorem and Binomial theorem for expressing sinnθ, cos nθ, interms

of powers of sin θ, cos θ and then using these expressions we express tan nθ, cot nθ in terms of

powers of tanθ, cot θ where n is a positive integer.

19.8.1

If n is a positive integer, θ is real number then

i)1 3 5

1 3 3 5 5C C Csin cos sin cos sin cos sin .......n n nn n n n− − −

θ θ θθ = θ − θ + θ +

wheren is an odd integer and

1 3 5

1 3 3 5 5C C Csin cos sin cos sin cos sin .......n n nn n n n− − −

θ θ θθ = θ − θ + θ

whenn is an even integer.

ii)2 4

2 2 4 4C Ccos cos cos sin cos sin .....n n nn n n− −θ = θ− θ + θ +

Where 'n' is an odd integer and

2 4

2 2 4 4 2C Ccos cos cos sin cos sin .....( 1) sin

nn n n nn n n− −θ = θ − θ + θ + − θ

Where n is an even integer.

174/14

19.8.2

If n is a positive integer, then

1 3 5 7

2 4 6

3 5 7C C C C

2 4 6C C C

Tan Tan Tan Tan ....Tan

1 Tan Tan Tan ....

n n n nn

n n n

θ − θ + θ − θ +θ =

− θ + θ − θ

providedcos 0nθ ≠

19.8.3 Note: If n is positive integer and sin 0nθ ≠ then we have cos

cotsin

nn

n

θθ =θ

2 4 6

1 3 5 7

2 4 6C C C

1 3 5 7C C C C

cot cot cot cot ....cot

cot cot cot cot ....

n n n n

n n n n

n n nn

n n n n

− − −

− − − −

θ − θ + θ− θ +θ =

θ − θ + θ − θ +

Example 1 : Expand i) cos6θ, ii) sin6θ, iii) tan6θ

Sol:i)2 4

2 2 4 4C Ccos cos cos sin cos sin .....n n nn n n− −

θθ = θ− θ θ + θ +

2 4 6

6 4 2 2 4 6C C Ccos6 cos 6 cos sin 6 cos sin 6 cosθ = θ− θ θ + θ θ − θ

6 4 2 2 4 6cos 15 cos sin 15 cos sin sin= θ− θ θ + θ θ − θ

ii)1 3

1 3 3C Csin cos sin cos sin .........n nn n n− −θ = θ θ − θ θ +

1 3 5

5 3 3 5C C Csin 6 6 cos sin 6 cos sin 6 cos sinθ = θ θ − θ θ + θ θ

5 3 3 56cos sin 20 cos sin 6cos sin= θ θ − θ θ + θ θ

iii) 1 3 5

2 4 6

3 5C C C

2 4 6C C C

Tan Tan Tan ....Tan

1 Tan Tan Tan ....

n n nn

n n n

θ − θ + θ −θ =

− θ + θ− θ + Hõ#∞Hõ

1 3 5

2 4 6

3 5C C C

2 4 6C C C

6 Tan 6 Tan 6 TanTan 6

1 6 Tan 6 Tan 6 Tan

θ − θ + θθ =

− θ + θ− θ

3 5

2 4 6

6Tan 20Tan 6Tan

1 15Tan 15Tan Tan

θ − θ + θ=− θ + θ − θ

174/15

Example 2 : Show that sin 6

sin

θθ = 32 cos5 θ − 32 cos3 θ + 6 cosθ 63 when sin 0θ ≠

Sol: sin 6θ = 6C1 cos5sinθ − 6C3

cos3 θ sin3 θ + 6C5 cosθ sin5θ

sin 6

sin

θθ = 6 cos5 θ − 20 cos3 θ sin2 θ + 6 cosθ sin4 θ

= 6 cos5 θ − 20 cos3 θ (1− cos2 θ) + 6 cosθ (1− cos2 θ)2

sin 6

sin

θθ = 32 cos5 θ − 32 cos3 θ + 6 cosθ .

19.9 Expressing sinn θ θ θ θ θ and cosnθθθθθ interms of sines and

cosines of multiples of θθθθθ

x = cosθ + i sin θ then 1

x = cos θ − i sin θ

1 12cos ; 2 sinx x i

x x+ = θ − = θ ...(1)

Letn be a positive integer, then by De-Moiver's theorem

1 12cos , 2 sinn n

n nx n x i nx x

+ = θ − = θ

1(cos sin ) cos sinn

n i n i nx

= θ + θ = θ + θ

1 12cos , 2 sinn n

n nx n x i nx x

+ = θ − = θ ...(2)

19.9.1

i) If 'n' is a positive integer then

2n−1 cosnθ = cos nθ + nC1 cos(n−2)θ + nC2

cos (n−4) θ + ...

19.9.2

Let n be a positive integer

i) If n is an even number then

2n−1 (−1)n/2sinnθ = cos nθ −nC1 cos (n−2)θ + nC2

(n−4) θ − ...

174/16

ii) If n is an odd number then

2n−1 (−1)n−1/2 sinnθ = sin nθ −nC1 sin (n−2)θ + nC2

(n−4) θ − ...

Example 3

Show that 27 cos8θ = cos8θ + 8 cos 6θ + 28 cos 4θ + 56 cos 2θ + 35

Sol : We have every even 'n'

2n−1 cosnθ = cos nθ +nC1 cos (n−2)θ + nC2

cos(n−4) θ + ... + 2

2

C( 1)

2 n

n

n−

putting n = 8 in the above expansion

27 cos8θ = cos8θ + 8C1 cos 6θ + 8C2

cos 4θ + 8C3 cos 2θ +

1

2 8C4

27 cos8θ = cos8θ + 8 cos 6θ + 28 cos 4θ + 56 cos 2θ + 35

Example 4

Show that 26 sin4θ cos3θ = cos 7θ − cos 5θ − 3cos 3θ + 3 cosθ.

Sol : x = cosθ + i sinθ then x + 1

x = 2 cos θ ; x −

1

x = 2i sin θ

1 12cos , 2 sinn n

n nx n x i nx x

+ = θ − = θ

4 34 3 1 1

(2 sin ) (2cos )i x xx x

θ θ = − +

3

22

1 1x x

xx

= − −

7 4 3 6 26 2

1 3 12 sin cos 3x x x

xx x

θ θ = − + − −

7 5 37 5 3

1 1 1 13 3x x x x

xx x x

= + − + − + + +

= 2 cos 7θ −2 sin 5θ − 3(2cos3θ) + 3(2 cos θ)

26 sin4θ cos3θ = cos 7θ − cos5θ − 3cos3 θ + 3 cos θ.

174/17


1. Show that 3sin 48cos 4cos

sin

θ = θ − θθ

2. Show that sin 7θ = 7sin θ − 56 sin3 θ + 112 sin5 θ − 64 sin7 θ

3. Show that 64 sin5θ cos2 θ = sin7 θ − 3 sin5 θ + sin3 θ + 5 sin θ

Letus sum up

1.1 3 5

1 3 3 5 5C C Csin cos sin cos sin cos sin ......n n nn n n n− − −

θ θ θθ = θ − θ + θ

2.2 4

2 2 4 4C Ccos cos cos sin cos sin ......n n nn n n− −θ = θ − θ θ + θ θ −

3. 1 3 5 7

2 4 6

3 5 7C C C C

2 4 6C C C

Tan Tan Tan Tan ....Tan

1 Tan Tan Tan ....

n n n nn

n n n

θ − θ + θ − θ +θ =

− θ + θ − θ +

4. If n is a positive integer then

(i) If 'n' is even then

2n−1 (−1)n/2 sinnθ = cos nθ −nC1 cos (n−2)θ + nC2

cos(n−4) θ − ...

(ii) If 'n' is odd then

2n−1

1

2( 1)n−

− sinnθ = sin nθ −nC1 sin (n−2)θ + nC2

sin(n−4) θ − ...

Supportive Websites



CALCULUS

20 Limit and Continuity21 Differentiation22 Differentiation of Trigonometric Functions23 Differentiation of Exponential and Logarithmic Functions24 Tangents and Normals25 Maxima and Minima26 Integration27 Definite Integrals28 Differential Equations

MODULE-V

MATHEMATICS 175

Notes

MODULE - VCalculus

Limit and Continuity

20

LIMIT AND CONTINUITY

Consider the function 2x 1f(x)

x 1

You can see that the function f(x) is not defined at x = 1 as x 1 is in the denominator. Take thevalue of x very nearly equal to but not equal to 1 as given in the tables below. In this casex 1 0 as x 1.

We can write 2 x 1 x 1x 1f x x 1

x 1 x 1, because x 1 0 and so division by

x 1 is possible.Table - 2

x f (x)1.9 2.91.8 2.81.7 2.71.6 2.61.5 2.5

: :: :

1.1 2.11.01 2.01

1.001 2.001: :: :

1.00001 2.00001

Table -1x f(x)

0.5 1.50.6 1.60.7 1.70.8 1.80.9 1.90.91 1.91

: :: :

0.99 1.99: :: :

0.9999 1.9999

In the above tables, you can see that as x gets closer to 1, the corresponding value of f (x) alsogets closer to 2.

MATHEMATICS

Notes

MODULE - VCalculus

176


However, in this case f(x) is not defined at x = 1. The idea can be expressed by saying that thelimiting value of f(x) is 2 when x approaches to 1.Let us consider another function f (x) =2x. Here, we are interested to see its behavior near thepoint 1 and at x = 1. We find that as x gets nearer to 1, the corresponding value of f (x) getscloser to 2 at x = 1 and the value of f (x) is also 2.So from the above findings, what more can we say about the behaviour of the function nearx = 2 and at x = 2 ?In this lesson we propose to study the behaviour of a function near and at a particular pointwhere the function may or may not be defined.

OBJECTIVES

After studying this lesson, you will be able to :● illustrate the notion of limit of a function through graphs and examples;● define and illustrate the left and right hand limits of a function y =f (x) at x = a;● define limit of a function y = f (x) at x = a;● state and use the basic theorems on limits;● establish the following on limits and apply the same to solve problems :

(i)n n

n 1x a

x alim na x ax a

(ii)x 0lim sinx 0 and

x 0lim cosx 1

(iii)x 0

s inxlim 1x

(iv)1x

x 0lim 1 x e

(vi)x

x 0

e 1lim 1x

(vii)x 0

log 1 xlim 1

x

● define and interprete geometrically the continuity of a function at a point;● define the continuity of a function in an interval;● determine the continuity or otherwise of a function at a point; and● state and use the theorems on continuity of functions with the help of examples.

EXPECTED BACKGROUND KNOWLEDGE● Concept of a function● Drawing the graph of a function● Concept of trigonometric function● Concepts of exponential and logarithmic functions

MATHEMATICS 177

Notes

MODULE - VCalculus


20.1 LIMIT OF A FUNCTION

In the introduction, we considered the function 2x 1f(x)

x 1. We have seen that as x approaches

l, f (x) approaches 2. In general, if a function f (x) approaches L when x approaches 'a', we saythat L is the limiting value of f (x)

Symbolically it is written as

x alim f x L

Now let us find the limiting value of the function 5x 3 when x approaches 0.

i.e. x 0lim 5x 3

For finding this limit, we assign values to x from left and also from right of 0.

x 0.1 0.01 0.001 0.0001..........5x 3 3.5 3.05 3.005 3.0005..........

x 0.1 0.01 0.001 0.0001..........5x 3 2.5 2.95 2.995 2.9995.........

It is clear from the above that the limit of 5x 3 as x 0 is -3

i.e., x 0lim 5x 3 3

This is illustrated graphically in the Fig. 20.1

Fig. 20.1

MATHEMATICS

Notes

MODULE - VCalculus

178


The method of finding limiting values of a function at a given point by putting the values of thevariable very close to that point may not always be convenient.We, therefore, need other methods for calculating the limits of a function as x (independentvariable) ends to a finite quantity, say a

Consider an example : Findx 3limf(x) , where

2x 9f(x)x 3

We can solve it by the method of substitution. Steps of which are as follows :

Step 1: We consider a value of x close to asay x = a + h, where h is a very small positivenumber. Clearly, as x a , h 0

Step 2 : Simplify f(x) f(a h)

Step 3 : Put h = 0 and get the requried result

For2x 9f(x)

x 3 we write x 3 h , so

that as x 3,h 0

Now f(x) f(3 h)

23 h 93 h 3

2h 6hh

= h + 6

x 3 h 0lim f x lim 6 h

As x 0 , h 0

Thus, x 3lim f x 6 0 6

by putting h = 0.

Remarks : It may be noted that f (3) is not defined, however, in this case the limit of thefunction f(x)asx 3 is 6.

Now we shall discuss other methods of finding limits of different types of functions.

Consider the example :

Findx 1limf(x),where

3

2x 1

, x 1f(x) x 11 , x 1

Here, for 3

2x 1x 1,f (x)x 1

2x 1 x x 1

x 1 x 1

MATHEMATICS 179

Notes

MODULE - VCalculus


It shows that if f (x) is of the form g(x)h(x)

, then we may be able to solve it by the method of

factors. In such case, we follow the following steps :

Step 1. Factorise g (x) and h (x) Sol.

3

2x 1f(x)x 1

2x 1 x x 1

x 1 x 1

( x 1, x 1 0∵ and as such canbe cancelled)

Step 2 : Simplify f (x)2x x 1f(x)x 1

Step 3 : Putting the value of x, we get the required limit.

Also f(1) 1(given)

In this case, x 1lim f ( x ) f (1)

3

2x 1

x 1 1 1 1 3lim1 1 2x 1

Thus, the limit of a function f (x) as x a may be different from the value of the functioin atx a .Now, we take an example which cannot be solved by the method of substitutions or method offactors.

Evaluatex 0

1 x 1 xlimx

Here, we do the following steps :Step 1. Rationalise the factor containing square root.Step 2. Simplify.Step 3. Put the value of x and get the required result.Solution :

1 x 1 x 1 x 1 x1 x 1 xx x 1 x 1 x

2 2(1 x) (1 x)

x 1 x 1 x

MATHEMATICS

Notes

MODULE - VCalculus

180


(1 x) (1 x)x 1 x 1 x

1 x 1 xx 1 x 1 x

2xx 1 x 1 x x 0, It can be cancelled∵

21 x 1 x

x 0 x 0

1 x 1 x 2lim lim

x 1 x 1 x

21 0 1 0

21 1

= 1

20.2 LEFT AND RIGHT HAND LIMITSYou have already seen that x a means x takes values which are very close to 'a', i.e. eitherthe value is greater than 'a' or less than 'a'.In case x takes only those values which are less than 'a' and very close to 'a' then we say x isapproaches 'a' from the left and we write it as x a .Similarly, if x takes values which aregreater than 'a' and very close to 'a' then we say x is approaching 'a' from the right and we writeit as x a .

Thus, if a function f (x) approaches a limit 1� , as x approaches 'a' from left, we say that the left

hand limit of f ( x ) a s x a is 1� .

We denote it by writing

1x alim f x � or 1

h 0lim f a h , h 0�

Similarly, if f (x) approaches the limit 2� , as x approaches 'a' from right we say, that the right

hand limit of f(x) as x a is 2� .

We denote it by writing

2x alim f x � or 2

h 0lim f a h , h 0�

Working RulesFinding the right hand limit i.e., Finding the left hand limit, i.e,

x alim f x

x alim f x

MATHEMATICS 181

Notes

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Put x a h Put x a h

Find h 0lim f a h Find h 0

lim f a h

Note : In both cases remember that h takes only positive values.

20.3 LIMIT OF FUNCTION y = f(x) AT x = aConsider an example :

Find 2x 1limf(x),where f(x) x 5x 3

Here2

h 0x 1lim f(x) lim 1 h 5 1 h 3

2h 0lim 1 2h h 5 5h 3

=1 + 5 + 3 = 9 .....(i)

and2

h 0x 1lim f(x) lim (1 h) 5(1 h) 3

2x 0lim 1 2h h 5 5h 3

1 5 3 9 .....(ii)

From (i) and (ii), x 1 x 1lim f(x) lim f(x)

Now consider another example :

Evaluate :x 3

| x 3 |limx 3

Here h 0x 3

| x 3 | |(3 h) 3 |lim limx 3 [(3 h) 3]

h 0

| h |limh

h 0

hlimh

(as h>0, so | h | = h)

=1 .....(iii)

and h 0x 3

| x 3 | |(3 h) 3 |lim limx 3 [(3 h) 3]

h 0

| h |limh

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182


h 0

hlimh

(as h > 0, so |-h | = h)

1 .....(iv)

From (iii) and (iv), x 3 x 3

| x 3 | | x 3 |lim limx 3 x 3

Thus, in the first example right hand limit = left hand limit whereas in the second example righthand limit left hand limit.

Hence the left hand and the right hand limits may not always be equal.

We may conclude that

2x 1lim x 5x 3 exists (which is equal to 9) and

x 3

| x 3 |limx 3

does not exist.

Note :

x ax a

x a

I lim f xlim f x

and lim f x

��

�

1x a

x a2x a

II lim f xlim f x doesnotexist.

and lim f x

�

�

IIIx alim f x or

x alim f x does not exist

x alim f x does not exist.

20.4 BASIC THEOREMS ON LIMITS

1.x a x alim cx c lim x, c being a constant.

To verify this, consider the function f x 5x.

We observe that in x 2lim 5x , 5 being a constant is not affected by the limit.

x 2 x 2lim5x 5 lim x

= 5 × 2 = 10

2.x alim g x h x p x ....

x a x a x alim g x lim h x lim p x ........

where g x ,h x ,p x ,.... are any function.

3.x a x a x alim f x g x lim f x lim g x

MATHEMATICS 183

Notes

MODULE - VCalculus


To verify this, consider 2f(x) 5x 2x 3

and g (x) = x + 2.

Then2

x 0 x 0lim f ( x ) lim 5x 2x 3

2x 0 x 0

5 lim x 2 lim x 3 3

x 0 x 0 x 0lim g(x) lim(x 2) lim x 2 2

2x 0 x 0lim(5x 2x 3) lim(x 2) 6 .....(i)

Again2

x 0 x 0lim[f(x) g(x)] lim[(5x 2x 3)(x 2)]

3 2x 0lim(5x 12x 7x 6)

=3 2

x 0 x 0 x 05 lim x 12 lim x 7 lim x 6

= 6 .....(ii)

From (i) and (ii),2 2

x 0 x 0 x 0lim[(5x 2x 3)(x 2)] lim(5x 2x 3) lim(x 2)

4.x a

x ax a

lim f xf xlimg x lim g x provided x a

lim g x 0

To verify this, consider the function2x 5x 6f(x)

x 2

we have2 2

x 1lim (x 5x 6) ( 1) 5 ( 1) 6

1 5 6

= 2

and x 1lim (x 2) 1 2

=1

2x 1

x 1

lim (x 5x 6) 2 2lim (x 2) 1 .....(i)

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184


Also

2

2 2

x 1 x 1

x 5x 6(x 5x 6) (x 3)(x 2) x 3x 2x 6lim lim

x 2 x 2 x(x 3) 2(x 3)(x 3)(x 2)

∵

x 1lim (x 3)

1 3 2 .....(ii) From (i) and (ii),

22

x 1x 1

x 1

lim x 5x 6x 5x 6limx 2 lim (x 2)

We have seen above that there are many ways that two given functions may be combined toform a new function. The limit of the combined function as x a can be calculated from thelimits of the given functions. To sum up, we state below some basic results on limits, which canbe used to find the limit of the functions combined with basic operations.

Ifx alimf (x ) � and

x alim g(x) m, then

(i) x a x alim kf(x) k limf (x ) k� where k is a constant.

(ii) x a x a x alim f(x) g(x) lim f ( x ) limg(x) m�

(iii) x a x a x alim f(x) g(x) lim f ( x ) limg(x) m�

(iv)x a

x a x ax a

limf (x )f(x)lim , provided limg(x) 0g(x) limg(x) m

�

The above results can be easily extended in case of more than two functions.

Example 20.1 Findx 1limf(x) , where

2x 1 , x 1f x x 11, x 1

Solution :2x 1

f(x)x 1

x 1 x 1x 1

MATHEMATICS 185

Notes

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(x 1) [ x 1]∵

x 1 x 1limf(x) lim(x 1)

= 1 + 1 = 2

Note : 2x 1

x 1is not defined at x=1. The value of

x 1limf(x) is independent of the value of f (x)

at x = 1.

Example 20.2 Evaluate : 3

x 2

x 8limx 2

.

Solution :3

x 2

x 8limx 2

2

x 2

(x 2)(x 2x 4)lim(x 2)

2x 2lim x 2x 4 [ x 2]∵

22 2 2 4= 12

Example 20.3 Evaluate : x 2

3 x 1lim .2 x

Solution : Rationalizing the numerator, we have

3 x 1 3 x 1 3 x 12 x 2 x 3 x 1

3 x 1(2 x) 3 x 1

2 x2 x 3 x 1

x 2 x 2

3 x 1 2 xlim lim

2 x 2 x 3 x 1

x 2

1lim

3 x 1 [ x 2]∵

13 2 1

11 1

12

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186


Example 20.4 Evaluate :x 3

12 x xlim6 x 3

.

Solution : Rationalizing the numerator as well as the denominator, we get

x 3 x 3

12 x x 12 x x 6 x 312 x xlim lim6 x 3 6 x 3 6 x 3 12 x x

2

x 3 x 3

12 x x 6 x 3lim lim6 x 9 12 x x

x 3 x 3

x 4 x 3 6 x 3lim lim

x 3 12 x x [ x 3]∵

=6(3 4) 76

Note : Whenever in a function, the limits of both numerator and denominator are zero, youshould simplify it in such a manner that the denominator of the resulting function is not zero.However, if the limit of the denominator is 0 and the limit of the numerator is non zero, thenthe limit of the function does not exist.

Let us consider the example given below :

Example 20.5 Find x 0

1limx

, if it exists.

Solution : We choose values of x that approach 0 from both the sides and tabulate the

correspondling values of 1x

.

x 0.1 .01 .001 .00011

10 100 1000 10000x

x 0.1 .01 .001 .00011

10 100 1000 10000x

We see that as x 0 , the corresponding values of 1x

are not getting close to any number..

Hence,x 0

1limx

does not exist. This is illustrated by the graph in Fig. 20.2

MATHEMATICS 187

Notes

MODULE - VCalculus


Fig. 20.2

Example 20.6 Evaluate : x 0lim | x | | x |

Solution : Since |x| has different values for x 0 and x<0, therefore we have to find out bothleft hand and right hand limits.

h 0x 0lim | x | | x | lim | 0 h | | (0 h ) |

h 0lim | h | | ( h ) |

= h 0 h 0limh h lim 2h 0 ...(i)

and h 0x 0lim | x | | x | lim | 0 h | | (0 h ) |

x 0 h 0lim h h lim 2h 0 ...(ii)

From (i) and (ii),

x 0 h 0lim | x | | x | lim | x | | x |

Thus , h 0lim | x | | x | 0

Note : We should remember that left hand and right hand limits are specially used when (a)the functions under consideration involve modulus function, and (b) function is defined bymore than one rule.

MATHEMATICS

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188


Example 20.7 Find the vlaue of 'a' so that

x 1limf(x) exist, where 3x 5 ,x 1

f(x)2x a,x 1

Solution :x 1x 1

lim f(x) lim 3x 5 f(x) 3x 5 f o r x 1∵

h 0lim 3 (1 h) 5

= 3 + 5 = 8 .....(i)

x 1x 1lim f(x) lim 2x a f(x) 2x a for x 1∵

h 0lim 2(1 h) a

= 2 + a .....(ii)

We are given that x 1limf(x) will exists provided

x 1 x 1lim lim f(x)

From (i) and (ii),2 + a = 8or, a = 6

Example 20.8 If a function f (x) is defined as1

x , 0 x2

1f(x) 0 , x

21x 1, x 12

Examine the existence of 1x2

limf(x) .

Solution : Here

1x , 0 x .....(i)

21

f(x) 0 , x2

1x 1 , x 1 .....(ii)2

h 01x2

1lim f(x) lim f h2

=x 0

1lim h2

1 1 1 1h and from(i),f h h

2 2 2 2∵

MATHEMATICS 189

Notes

MODULE - VCalculus


1 102 2

.....(iii)

h 01x2

1lim f(x) lim f h2

h 0

1lim h 12

1 1 1 1h and from(ii),f h h 12 2 2 2

∵

=1 12

12

.....(iv)

From (iii) and (iv), left hand limit right hand limit

1x2

limf(x) does not exist.

CHECK YOUR PROGRESS 20.11. Evaluate each of the following limits :

(a) x 2lim 2(x 3) 7 (b) 2

x 0lim x 3x 7 (c)

2x 1lim (x 3) 16

(d) 2x 1lim (x 1) 2 (e) 3

x 0lim (2x 1) 5 (f) x 1

lim 3x 1 x 1

2. Find the limits of each of the following functions :

(a)x 5

x 5limx 2

(b)x 1

x 2limx 1

(c)x 1

3x 5limx 10

(d)x 0

px qlimax b

(e)2

x 3

x 9limx 3

(f)2

x 5

x 25limx 5

(g)2

2x 2

x x 2limx 3x 2

(h)2

1x3

9x 1lim3x 1

3. Evaluate each of the following limits:

(a)3

x 1

x 1limx 1

(b)3

2x 0

x 7xlimx 2x

(c)4

x 1

x 1limx 1

(d) 2x 1

1 2limx 1 x 1

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190


4. Evaluate each of the following limits :

(a)x 0

4 x 4 xlimx

(b)x 0

2 x 2limx

(c)x 3

3 x 6limx 3

(d)x 0

xlim1 x 1 (e)

x 2

3x 2 xlim

2 6 x

5. (a) Find x 0

2limx

, if it exists. (b) Find x 2

1limx 2

, if it exists.

6. Find the values of the limits given below :

(a)x 0

xlim5 | x | (b)

x 2

1lim| x 2 | (c)

x 2

1lim| x 2 |

(d) Show that x 5

| x 5 |limx 5

does not exist.

7. (a) Find the left hand and right hand limits of the function

2x 3, x 1f(x)

3x 5,x 1 as x 1

(b) If 2

x 1

x , x 1f(x) ,findlimf(x)1,x 1

(c) Find x 4limf(x) if it exists, given that 4x 3,x 4

f(x)3x 7,x 4

8. Find the value of 'a' such that x 2limf(x) exists,where ax 5, x 2

f(x)x 1, x 2

9. Let2

x,x 1f(x) 1,x 1

x ,x 1

Establish the existence of x 1limf(x) .

10. Find x 2limf(x) if it exists, where

x 1,x 2f(x) 1,x 2

x 1,x 2

20.5 FINDING LIMITS OF SOME OF THE IMPORTANTFUNCTIONS

(i) Prove that n n

n 1x a

x alim nax a

where n is a positive integer..

MATHEMATICS 191

Notes

MODULE - VCalculus


Proof :n nn n

x a h 0

a h ax alim limx a a h a

n n 1 n 2 2 n n

h 0

n n 1a n a h a h ..... h a

2!lim

h

n 1 n 2 n 1

h 0

n n 1h n a a h ..... h

2!lim

h

n 1 n 2 n 1h 0

n n 1lim n a a h ..... h

2!n 1n a 0 0 ..... 0n 1n a

n nn 1

x a

x alim n ax a

Note : However, the result is true for all n

(ii) Prove that (a) x 0lim sinx 0 and (b) x 0

lim cosx 1

Proof : Consider a unit circle with centre B, in which C is a right angle and ABC = xradians.

Now sin x = A C and cos x = BC

As x decreases, A goes on coming nearer and nearer to C.

i.e., when x 0,A C

or when x 0,AC 0

and BC AB,i.e.,BC 1

When x 0 sinx 0 and cosx 1

Thus we have

x 0 x 0lim sinx 0 and lim cosx 1

(iii) Prove thatx 0

s inxlim 1x

Proof : Draw a circle of radius 1 unit and with centre at the origin O. Let B (1,0) be a point onthe circle. Let A be any other point on the circle. Draw AC OX .

Fig. 20.3

MATHEMATICS

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192


Let AOX x radians, where 0<x<2

Draw a tangent to the circle at B meeting OA producedat D. Then BD OX .

Area of AOC area of sector OBA area ofOBD .

or 21 1 1OC AC x(1) OB BD2 2 2

21 1area of triangle base×height and area of sector= r

2 2∵

1 1 1cosxsinx x 1 tanx2 2 2

OC AC BDcosx ,sinx and tanx ,OA 1 OB

OA OA OB∵

i.e.,x tanxcosx

sinx sinx[Dividing throughout by

12

sin x]

orx 1cosx

sinx cosx

or1 sinx cosx

cosx x

i.e.,sinx 1cosx

x cosxTaking limit as x 0 , we get

x 0 x 0 x 0

sinx 1lim cosx lim limx cosx

orx 0

s inx1 lim 1x x 0 x 0

1 1lim cosx 1 and lim 1

cosx 1∵

Thus,x 0

s inxlim 1x

Note : In the above results, it should be kept in mind that the angle x must be expressed inradians.

(iv) Prove that

1x

x 0lim 1 x e

Proof : By Binomial theorem, when | x | 1, we get

Fig. 20.4

MATHEMATICS 193

Notes

MODULE - VCalculus


1x 2 3

1 1 1 1 11 1 21 x x x x x1 x 1 x x x ..........x 2! 3!

1 x 1 x 1 2x1 1 ..........

2! 3!1x

x 0 x 0

1 x 1 2x1 xlim 1 x lim 1 1 ..........2! 3!

1 11 1 ..........

2! 3!= e (By definition)

Thus1x

x 0lim 1 x e

(v) Prove that

1 / x

x 0 x 0 x 0

log 1 x 1lim lim log 1 x limlog 1 xx x

= log e

1x

x 0Using lim 1 x e

= 1

(vi) Prove that x

x 0

e 1lim 1x

Proof : We know that2 3

x x xe 1 x ..........2! 3!

2 3x x xe 1 1 x .......... 1

2! 3!

2 3x xx ..........2! 3!

2 3

xx xx ..........2! 3!e 1

x x[Dividing throughout by x]

2x xx 1 ..........2! 3!

x

MATHEMATICS

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194


2x x1 ..........2! 3!

x 2

x 0 x 0

e 1 x xlim lim 1 ..........x 2! 3!

1 0 0 ........ 1

Thus,x

x 0

e 1lim 1x

Example 20.9 Find the value ofx x

x 0

e elimx

Solution : We know thatx

x 0

e 1lim 1x

.....(i)

Putting x x in (i), we getx

x 0

e 1lim 1x

.....(ii)

Given limit can be written asx x

x 0

e 1 1 elimx

[Adding (i) and (ii)]

x x

x 0

e 1 1 elimx x

x x

x 0

e 1 e 1limx x

x x

x 0 x 0

e 1 e 1lim limx x

= 1+1 [ Using (i) and (ii)]= 2

Thusx x

x 0

e elim 2x

Example 20.10 Evaluate :x

x 1

e elimx 1

.

Solution : Put x = 1 + h, where h 0x 1 h

x 1 h 0

e e e elim limx 1 h

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Notes

MODULE - VCalculus


1 h

h 0

e e elimh

h

h 0

e(e 1)limh

h

h 0

e 1elimh

= e × 1 = e.

Thusx

x 1

e elim ex 1

Example 20.11 Evaluate : x 0

sin3xlimx

.

Solution :x 0 x 0

sin3x sin3xlim lim 3x 3x

[Multiplying and dividing by 3]

3x 0

sin3x3 lim3x

[ when x 0,3x 0]∵

x 0

s i n x3.1 lim 1

x∵

= 3

Thus,x 0

sin3xlim 3x

Example 20.12 Evaluate2x 0

1 cosxlim2x

.

Solution : 222

2 2x 0 x 02

x cos2x 1 2sin x,2sin1 c o s x 2lim lim 1 cos2x 2sin x2x 2x xor 1 cosx 2sin

2

∵

=

2

x 0

xsin2lim x22

[Multiplying and dividing the denominator by 2]

2

x 02

xsin1 2lim x42

MATHEMATICS

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196


1 114 4

2x 0

1 cosx 1lim42x


1 c o s 4lim1 c o s 6

Solution : 2

20 0

1 cos4 2sin 2lim lim1 cos6 2sin 3

=2 2

0

sin2 3 1lim 22 sin3 3

2 2

20

s in2 3 4lim2 s in3 9

2

2 0 3 0

4 s in2 3lim lim9 2 s in3

4 41 19 9

Example 20.14 Find the value of 2

x2

1 cos2xlim2x

.

Solution : Put x h2 ∵ when x , h 0

22x 2h

2 2h 0x2

1 cos2 h1 cos2x 2lim lim

[ ( 2h)]2x

2h 0

1 cos( 2h)lim4h

2h 0

1 cos2hlim4h

2

2h 0

2sin hlim4h

2

h 0

1 sin hlim2 h

MATHEMATICS 197

Notes

MODULE - VCalculus


1 112 2

2x

2

1 cos2x 1lim22x

Example 20.15 Evaluate x 0

s i n a xlimt a n b x

Solution : x 0 x 0

s i n a x as i n a x axlim lim tanbxt a n b x bbx

x 0

x 0

s i n a xlima axt a n b xb lim

bx

a 1b 1

ab

x 0

sinax alimtanbx b

CHECK YOUR PROGRESS 20.21. Evaluate each of the following :

(a)2x

x 0

e 1limx

(b)x x

x xx 0

e elime e

2. Find the value of each of the following :

(a)x 1

x 1

e elimx 1

(b)x

x 1

e elimx 1

3. Evaluate the following :

(a)x 0

sin4xlim2x

(b) 2

2x 0

sinxlim5x

(c)2

x 0

sinxlimx

(d) x 0

sinaxlimsinbx

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198

Limit and Continuity4. Evaluate each of the following :

(a) 2x 0

1 cosxlimx (b) x 0

1 cos8xlimx (c) 3x 0

sin2x(1 cos2x)limx

(d) 2x 0

1 cos2xlim3tan x

5. Find the values of the following :

(a) x 0

1 cosaxlim1 cosbx (b)

3

x 0

x cotxlim1 cos x (c)

x 0

cosec x c o t xlimx


(a) x

sinxlimx (b)

x 1

cos x2lim

1 x(c)

x2

lim sec x t a n x

7. Evaluate the following :

(a)x 0

sin5xlimtan3x

(b)0

t a n 7lims i n 4

(c)x 0

sin2x tan3xlim4x tan5x

20.5 CONTINUITY OF A FUNCTION AT A POINT

Fig. 20.5

Let us observe the above graphs of a function.

We can draw the graph (iv) without lifting the pencil but in case of graphs (i), (ii) and (iii), thepencil has to be lifted to draw the whole graph.

In case of (iv), we say that the function is continuous at x = a. In other three cases, the functionis not continuous at x = a. i.e., they are discontinuous at x = a.

In case (i), the limit of the function does not exist at x = a.

In case (ii), the limit exists but the function is not defined at x = a.

In case (iii), the limit exists, but is not equal to value of the function at x = a.

In case (iv), the limit exists and is equal to value of the function at x = a.

MATHEMATICS 199

Notes

MODULE - VCalculus


Example 20.16 Examine the continuity of the function f (x) = x a at x = a.

Solution : x a h 0lim f ( x ) l imf(a h)

h 0lim[(a h) a]

= 0 .....(i)Also f(a) a a 0 .....(ii)From (i) and (ii),

x alim f ( x ) f(a)

Thus f (x) is continuous at x = a.

Example 20.17 Show that f (x) = c is continuous.

Solution : The domain of constant function c is R.Let 'a' be any arbitrary real number.

x alim f ( x ) candf(a) c

x alim f ( x ) f(a)

f (x) is continuous at x = a. But 'a' is arbitrary. Hence f(x) = c is a constant function.

Example 20.18 Show that f x cx d is a continuous function.

Solution : The domain of linear function f x cx d is R; and let 'a' be any arbitrary realnumber.

x a h 0lim f ( x ) l imf(a h)

h alim[c (a h) d]

ca d .....(i)

Also f a ca d .....(ii)

From (i) and (ii), x alim f ( x ) f(a)

f (x) is continuous at x = aand since a is any arbitrary , f (x) is a continuous function.

Example 20.19 Prove that f x sin x is a continuous function.Solution : Let f (x) = sin xThe domain of sin x is R. let 'a' be any arbitrary real number.

x a h 0limf(x) limf(a h)

= h 0limsin(a h)

= h 0lim [sina.cosh cosa.sinh]

MATHEMATICS

Notes

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200


= h 0 h 0sinalimcosh cosa limsinh

x a x alim kf(x) k limf(x)where k is a constant∵

sina 1 cosa 0x 0 x 0l imsinx 0 and lim cos x 1∵

= sin a .....(i)Also f (a) = sin a .....(ii)

From (i) and (ii), x alimf(x) f(a)

sin x is continuous at x = a∵sin x is continuous at x = a and 'a' is an aribitary point.

Therefore, f x sinx is continuous.

Example 20.20 Given that the function f (x) is continuous at x = 1. Find a when,

f x ax 5 and f (1) = 4.

Solution : x 1 h 0limf(x) lim(1 h)

= h 0lima(1 h) 5

= a + 5 .....(i)

Also f (1) = 4 .....(ii)

As f (x) is continuous at x = 1, therefore x 1limf(x) f(1)

From (1) and (ii), a +5 = 4

or a 4 5 or a 1

Definition :1. A function f (x) is said to be continuous in an open inteval ]a,b[ if it is continuous at every

point of ]a,b[*.2. A function f (x) is said to be continuous in the closed interval [a,b] if it is continuous at

every point of the open interval ]a,b[ and is continuous at the point a from the right andcontinuous at b from the left.

i.e.x alim f x f a

andx blim f x f b

* In the open interval ]a,b[ we do not consider the end points a and b.

MATHEMATICS 201

Notes

MODULE - VCalculus


Example 20.21 Prove that tan x is continuous when 0 x2

Solution : Let f (x) = tan x

The domain of tan x is R (2n 1) , n I2

Let a R (2n 1)2

, be arbitrary..

x a h 0lim f ( x ) l imf(a h)

h 0lim tan(a h)

h 0

sin(a h)limcos(a h)

h 0

sinacosh cosasinhlimcosacosh sinasinh

h 0 h 0

h 0 h 0

s i n a l i m c o s h cosa lim sin h

cosa l im cos h s i n a l i m s i n h

s ina 1 cosa 0cosa 1 s ina 0

=sinacosa

= tan a .....(i) a Domain of tan x, cos a 0∵Also f (a) = tan a .....(ii)

From (i) and (ii), h al imf(x) f(a)

f(x) is continuous at x = a. But 'a' is arbitrary..

tan x is continuous for all x in the interval 0 x2

.


1. Examine the continuity of the functions given below :(a) f(x) x 5atx 2 (b) f(x) 2x 7atx 0

(c)5f(x) x 7atx 33

(d) f(x) px qa tx q

2. Show that f (x)=2a+3b is continuous, where a and b are constants.3. Show that 5 x + 7 is a continuous function

4. (a) Show that cos x is a continuous function.

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202

Limit and Continuity(b) Show that cot x is continuous at all points of its domain.

5. Find the value of the constants in the functions given below :

(a) f(x) px 5andf(2) 1 such that f (x) is continuous at x = 2.

(b) f(x) a 5xandf(0) 4 such that f (x) is continuous at x = 0.

(c)2f(x) 2x 3bandf( 2)3

such that f(x) is continuous at x 2 .

20.7 CONTINUITY OR OTHERWISE OF A FUNCTION AT APOINT

So far, we have considered only those functions which are continuous. Now we shall discusssome examples of functions which may or may not be continuous.

Example 20.22 Show that the function xf(x) e is a continuous function.

Solution : Domain of xe is R. Let a R . where 'a ' is arbitrary..

x a h 0limf(x) limf(a h) , where h is a very small number..

a hh 0l ime

a hh 0lime e

a hh 0

e lim e

ae 1 .....(i)

ae .....(ii)

Also f (a) ae


f (x) is continuous at x = a

Since a is arbitary, xe is a continuous function.

Example 20.23 By means of graph discuss the continuity of the function 2x 1f x

x 1.

Solution : The grah of the function is shown in the adjoining figure. The function is discontinuousas there is a gap in the graph at x = 1.

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Fig. 20.7

Fig. 20.6


1. (a) Show that 5xf x e is a continuous function.

(b) Show that 2x

3f x e is a continuous function.

(c) Show that 3x 2f x e is a continuous function.

(d) Show that 2x 5f x e is a continuous function.

2. By means of graph, examine the continuity of each of the following functions :

(a) f (x) = x+1. (b)x 2f xx 2

(c)2x 9f x

x 3(d)

2x 16f xx 4

20.6 PROPERTIES OF CONTINUOUS FUNCTIONS(i) Consider the function f (x) = 4. Graph of the function

f (x) = 4 is shown in the Fig. 20.7. From the graph,we see that the function is continuous. In general,all constant functions are continuous.

(ii) If a function is continuous then the constant multipleof that function is also continuous.

Consider the function 7f x x2

. We know that x

is a constant function. Let 'a' be an arbitrary realnumber.

x a h 0limf(x) limf(a h)

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204


h 0

7lim (a h)2

7 a2

.....(i)

Also7f(a) a2

.....(ii)

From (i) and (ii),

x alimf(x) f(a)

7f x x2

is continuous at x = a.

As72

is constant, and x is continuous function at x = a, 72

x is also a continuous function at x = a.

(iii) Consider the function 2f(x) x 2x . We know that the function 2x and 2x arecontinuous.

Now x a h 0limf(x) limf(a h)

2

h 0lim a h 2 a h

2 2h 0lim a 2ah h 2a 2ah

2a 2a .....(i)

Also 2f(a) a 2a .....(ii)


f(x) is continuous at x = a.

Thus we can say that if 2x and 2x are two continuous functions at x = a then 2x 2x is also

continuous at x = a.

(iv) Consider the function 2f(x) x 1 x 2 . We know that 2x 1 and x 2 are

two continuous functions.

Also 2f(x) x 1 x 2

3 2x 2x x 2As 3 2x ,2x , x and 2 are continuous functions, therefore.

3 2x 2x x 2 is also a continuous function.

We can say that if 2x 1 and (x+2) are two continuous functions then 2x 1 (x 2)

is also a continuous function.

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(v) Consider the function 2x 4f(x)

x 2 at x = 2. We know that 2x 4 is continuous at

x = 2. Also ( x + 2) is continuous at x = 2.

Again2

x 2 x 2

x 2 x 2x 4lim limx 2 x 2

x 2lim x 2

2 2 0

Also2(2) 4f(2)

2 2

0 04

x 2limf(x) f(2). Thus f (x) is continuous at x = 2.

If 2x 4 and x + 2 are two continuous functions at x = 2, then 2x 4

x 2 is also continuous.

(vi) Consider the function f(x) | x 2 | . The function can be written as

(x 2),x 2f(x)

(x 2),x 2

h 0x 2lim f(x) lim f ( 2 h) , h > 0

h 0lim (2 h) 2

2 2 0

h 0x 2lim f(x) l imf(2 h) , h > 0 ......(i)

x 2lim (2 h) 2

2 2 0 .....(ii)

Also f(2) (2 2) 0 .....(iii)

From (i), (ii) and (iii), x 2limf(x) f(2)

Thus, | x 2 | is continuous at x = 2.

After considering the above results, we state below some properties of continuous functions.

If f (x) and g (x) are two functions which are continuous at a point x = a, then

(i) C f (x) is continuous at x = a, where C is a constant.

(ii) f(x) g(x) is continuous at x = a.

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206


(iii) f x g x is continuous at x = a.

(iv) f (x)/g (x) is continuous at x = a, provided g (a) 0.

(v) |f(x)| is continuous at x = a.

Note : Every constant function is continuous.

20.9 IMPORTANT RESULTS ON CONTINUITYBy using the properties mentioned above, we shall now discuss some important results oncontinuity.

(i) Consider the function f(x) px q,x R (i)The domain of this functions is the set of real numbers. Let a be any arbitary real number.Taking limit of both sides of (i), we have

x a x alim f ( x ) lim px q pa q value of p x +q at x = a.

px +q is continuous at x = a.

Similarly, if we consider 2f(x) 5x 2x 3 , we can show that it is a continuous function.

In general 2 n 1 n0 1 2 n 1 nf(x) a a x a x ... a x a x

where 0 1 2 na ,a ,a .....a are constants and n is a non-negative integer,,

we can show that 2 n0 1 2 na ,a x,a x ,.....a x are all continuos at a point x = c (where c is any

real number) and by property (ii), their sum is also continuous at x = c. f (x) is continuous at any point c.

Hence every polynomial function is continuous at every point.

(ii) Consider a function x 1 x 3f(x) ,f(x)

x 5is not defined when x 5 0 i.e, at x = 5.

Since (x + 1) and (x + 3) are both continuous, we can say that (x + 1) (x + 3) is alsocontinuous. [Using property iii]

Denominator of the function f (x), i.e., (x 5) is also continuous.

Using the property (iv), we can say that the function x 1 x 3

(x 5)is continuous at all

points except at x = 5.

In general if p(x)f(x)q(x)

, where p (x) and q (x) are polynomial functions and q (x) 0,

then f (x) is continuous if p (x) and q (x) both are continuous.

Example 20.24 Examine the continuity of the following function at x = 2.

3x 2 forx 2f(x)

x 2 forx 2

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Solution: Since f(x) is defined as the polynomial function 3x−2 on the left hand side of thepoint x = 2 and by another polynomial function x +2 on the right hand side of x = 2, we shallfind the left hand limit and right hand limit of the function at x = 2 separately.

Fig. 20.8

Left hand limit = x 2lim f (x)

−→

x 2lim(3x 2)→

= −

= 3 × 2 − 2 = 4

Right hand limit at x = 2;

x 2x 2lim f (x) lim(x 2) 4

+ →→= + =

Since the left hand lirmt and the right hand limit at x = 2 are equal, the limit 0 the function f (x)

exists at x = 2 and is equal to 4 i.e., x 2lim f (x) 4→

= .

Also f(x) is defined by (x +2) at x = 2

f (2) = 2 + 2 = 4

Thus x 2lim f (x) 2→

=

Hence (x) is continuous at x = 2.

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Example 20.25

(i) Draw the graph of f(x) = |x|.

(ii) Discusss the continuity of f(x) at x = 0.

Solution: We know that for x > 0, | x | = x and for x < 0, | x | = −x. Hence f(x) can be writtenas.

x, x 0f (x) | x |

x, x 0

− <= = ≥

(i) The graph of the function is given in Fig 20.9

Fig. 20.9

(ii) Left hand limit = x 0lim f (x)

−→

x 0lim ( x) 0→

= − =

Right hand limitx 0lim f (x)

+→=

x 0lim x 0→

= =

Thus,x 0lim f (x) 0→

=

A1so, f(x) = 0

∴x 0lim f (x) f (0)→

=

Hence the function f (x) is continuous at x = 0.

Example 20.26 Examine the continuity of f (x) = |x − b| at x = b.

Solution: We have f(x) = | x − b |. This function can be written as

(x b), x bf (x)

(x b), x b

− − <= − ≥

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Notes


Left hand limith 0x b

lim f (x) lim f (b h)− →→

= = −

h 0lim[ (b h b)]→

= − − −

h 0lim h 0→

= = ...(i)

Righthandlimith 0x b

lim f (x) lim f (b h)+ →→

= = +

h 0lim f (b h)→

= +

h 0lim h 0→

= = ...(ii)

Also, f(b) = b − b = 0

From (i), (ii) and (iii), x blim f (x) f (b)→

= ...(iii)

Thus, f(x) is continuous at x = b.

Example 20.27 If sin 2x

, x 0f (x) x

2, x 0

≠= =

find whether f(x) is continuous at x = 0 or not.

Solution: Here sin 2x

, x 0f (x) x

2, x 0

≠= =

Left hand limit = x 0

sin 2xlim

x−→

h 0

sin 2(0 h)lim

0 h→

−=−

h 0

sin 2hlim

h→

−=−

h 0

sin 2h 2lim

h 1→

= × = 1 ×2 = 2 ...(i)

Right hand limitx 0

sin 2xlim

x+→=

h 0

sin 2(0 h)lim

0 h→

+=+

h 0

sin 2h 2lim

2h 1→= ×

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Notes


= l × 2 = 2

Also f(0) = 2

From (i) to (iii),

x 0lim f (x) 2 f (0)→

= =

Hence f(x) is continuous at x = 0.

Example 20.28 If 2x 1

f (x)x 1

−=−

for x ≠ 1 and f(x) = 2 when x = 1, show that the function

f(x) is continuous at x = 1.

Solution : Here

2x 1, x 1

f (x) x 12 , x 1

− ≠= − =

Left hand limitx 1lim f (x)

−→

h 0lim f (1 h)→

= −

2

h 0

(1 h) 1lim

(1 h) 1→

− −=− −

2

h 0

1 2h h 1lim

h→

− + −=−

h 0

h(h 2)lim

h→

−=−

h 0lim (h 2)→

= − −

= 2

Right hand limitx 1lim f (x)

+→= ...(i)

h 0lim f (1 h)→

= +

2

h 0

(1 h) 1lim

(1 h) 1→

+ −=+ −

2

h 0

1 2h h 1lim

h→

+ + −=

h 0

h(h 2)lim

h→

+=

= 2 ...(ii)Also f(1) = 2 (Given) ...(iii)∴ From (i) to (iii),

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Notes


x 1limf (x) f (1)

→=

Thus f(x) is continuous at x = 1.

Example 20.29 Find whether f (x) is continuous at x = 0 or not, where

x, x 0

f (x) x2, x 0

≠= =

Solution : x 0 x 0

xlim f (x) lim

x→ →=

x 0lim1 1→

= =

and f(0) = 2

∴x 0lim f (x) f (0)→

≠ Fig. 20.10

Hence f (x) is not continuous at x = 0. The graph of the function is given in Fig. 20.10.

Clearly, the point (0,1) does not lie on the graph. Therefore, the function is discontinuous atx = 0.

Signum Function : The function f (x) = sgn(x) (read as signum x) is defined as

1, x 0

f (x) 0, x 0

1, x 0

− <= = >

Find the left hand limit and right hand limit of the function from its graph given below:

Fig. 20.11

From the graph, we see that as x → 0+, f(x) → 1 and as (x) → 0−, f(x) →−1

Hence,x 0 x 0lim f (x) 1, lim f (x) 1

+ −→ →= = −

As these limits are not equal, x 0lim f (x)→

does not exist. Hence f(x) is discontinuous at x =0.

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Greatest Integer Function: Let us consider the function f(x) = [x] where [x] denotes thegreatest integerles than or equal to x. Find whether f (x) is continuous at

(i) x = 1

2(ii) x =1

To solve this, let us take some arbitrary values ofx say 1.3, 0.2, −0.2 ..... By the defmitionofgreatest integer function,

In general :

for −3 < x < −2 , [x] = −3

for −2 < x < −1, [x] = −2

for −1 < x < 0 , [x] = −1

for 0 < x < l, [x] = 0

for 1 < x < 2, [x] = 1 and so on.

The graph of the function f(x) = [x] is given in Fig. 20.12

(i) From graph

1 1x x

2 2

lim f (x) 0, lim f (x) 0,− +

→ →

= =

∴ 1x

2

lim f (x) 0→

=

Also 1

f [0.5] 02

= =

Thus1

x2

1lim f (x) f

2→

=

Hence f(x) is continuous at

1x

2=

(ii)x 1 x 1lim f (x) 0, lim f (x) 1

− +→ →= = Fig. 20.12

Thusx 1limf (x)

→, does not exist.

Hence, f (x) is discontinuous at x = 1.

Note : The function f(x) = [x] is also known as Step Function.

Example 20.30 At what points is the function x 1

(x 4)(x 5)

−+ −

continuous ?

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Notes


Solution: Here x 1

f (x)(x 4)(x 5)

−=+ −

The function in the numerator i.e., x −1 is continuous. lne function in the demoninator is (x + 4)(x − 5) which is also continuous.

But f(x) is notdefmed attheyoints − 4 and 5.

∴ The function f(x) is continuous at all points except − 4 and 5 at which it is not defined.

In other words, f (x) is continuous at all points of its domain.

Example 20.31 Find the points of discountinuity of the function 2

2

x 2x 5f (x) .

x 8x 12

+ +=− +

Solution: Here f (x) is a rational function.

Denominator = x2 − 8x + 12

= (x − 2)(x − 6) is zero at x = 2 and x = 6

∴ f(x) is not defined at x = 2 and x = 6.

Also we know that a rational function is continuous at all points of its domain.

∴ f(x) is continuous for all values ofx except x = 2 and x = 6.


1. (a) If f(x) = 2x + 1, when x1 and f(x)=3 when x = 1, show that the function f(x)continuous at x = 1.

(b) If 4x 3, x 2

f (x) ,3x 5, x 2

+ ≠= + =

find whether the function f is continuous at x = 2.

(c) Determine whether f (x) is continuous at x = 2, where

4x 3, x 2f (x)

8 x, x 2

+ ≤= − >

(d) Examine the continuity of f (x) at x = 1, where

2x , x 1f (x)

x 5, x 1

≤= + >

(e) Determine the values of k so that the function

2kx , x 2f (x)

3, x 2

≤= >

is continuous at x = 2.

2. Examine the continuity of the following functions :

(a) f(x) = |x−2| at x = 2 (b) f(x) = |x + 5| at x = −5

(c) f(x) =| a − x| at x = a

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Notes


(d)| x 2 |

, x 2f (x) x 2

1, x 2

− ≠= − =

at x = 2

(e)| x a |

, x af (x) x a

1, x a

− ≠= − =

at x = a

3. (a ) If sin 4x, x 0

f (x)2, x 0

≠= =

at x = 0

(b) If

sin 7x, x 0

f (x) x7, x 0

≠= =

at x = 0

(c) For what value of a is the function

sin 5x, x 0

f (x) 3xa, x 0

≠= =

continuous at x = 0 ?

4. (a) Show that the function f (x) is continuous at x = 2, where

2x x 2, for x 0

f (x) x 23, for x 2

− − ≠= − =

(b) Test the continuity of the function f(x) at x = 1, where

2x 4x 3, for x 1

f (x) x 12, for x 1

− + ≠= − − =

( c) For what value of k is the following function continuous at x = 1 ?

2x 1, when x 1

f (x) x 1k, when x 1

− ≠= − =

(d) Discuss the continuity of the function f(x) at x = 2, when

2x 4, for x 2

f (x) x 27, x 2

− ≠= − =

5. (a) If | x |

, x 0f (x) ,x

0, x 0

≠= =

find hether f is continuous at x = 0.

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Limit and Continuity(b) Test the continuity of the function f (x) at the origin.

wherex , x 0

f(x) | x |1, x 0

6. Find whether the function f (x)=[x] is continuous at

(a)4x3

(b) x = 3 (c) x 1 (d)2x3

7. At what points is the function f (x) continuous in each of the following cases ?

(a) x 2f(x)x 1 x 4

(b) x 5f(x)x 2 x 3

(c) 2

x 3f(x)x 5x 6

(d)2

2x 2x 5f(x)x 8x 16

NT� If a function f(x) approaches l when x approches a, we say that l is the limit of f (x).

Symbolically, it is written as

x alimf(x) �

� If x alimf(x) � and

x alimg(x) m,then

(i) x a x alimkf(x) k lim f ( x ) k�

(ii)x a x a x alim f(x) g(x) limf(x) limg(x) m�

(iii) x a x a x alim f(x)g(x) limf(x)limg(x) m�

(iv) x ax a

x a

limf(x)f(x)limg(x) limg(x) m

� , provided x alimg(x) 0

� LIMIT OF IMPORTANT FUNCTIONS

(i)n n

n 1x a

x alim nax a

(ii) x 0limsinx 0

(iii) x 0lim cosx 1 (iv)

x 0

sinxlim 1x

(v)1x

x 0lim 1 x e (vi)

x 0

log 1 xlim 1

x

(vii)x

x 0

e 1lim 1x

LET US SUM UP

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216



TERMINAL EXERCISE

Evaluate the following limits :

1. x 1l im5 2.

x 0lim 2

3.5

6 3x 1

4x 9x 7lim3x x 1

4.2

3 2x 2

x 2xlimx x 2x

5.4 4

x 0

(x k) xlimk(k 2x) 6.

x 0

1 x 1 xlimx

7. 2x 1

1 2limx 1 x 1

8.x 1

(2x 3) x 1lim

(2x 3)(x 1)

9.2

x 2

x 4limx 2 3x 2

10. 2x 1

1 2limx 1 x 1

11.x

sinxlimx

12.2 2

2 2x a

x (a 1)x alimx a

Find the left hand and right hand limits of the following functions :

13. 2x 3if x 1f(x) asx 1

3x 5 if x 114.

2x 1f(x) asx 1| x 1|

Evaluate the following limits :

15.x 1

| x 1|limx 1 16.

x 2

| x 2 |limx 2 17.

x 2

x 2lim| x 2 |

18. If 2(x 2) 4f(x)

x, prove that

x 0limf(x) 4 though f (0) is not defined.

19. Find k so that x 2limf(x) may exist where

5x 2,x 2f(x)

2x k,x 2

20. Evaluate x 0

sin7xlim2x

21. Evauate x x

2x 0

e e 2lim

x


MATHEMATICS 217

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22. Evaluate 2x 0

1 cos3xlimx

23.Find the value of x 0

sin2x 3xlim2x sin3x

24. Evaluate x 1

xlim(1 x)tan2

25. Evaluate 0

sin5limtan8

Examine the continuity of the following :

26.1 3x if x 1

f(x)2 i f x 1

at x 1

27.

1 1x,0 x

x 21 1f(x) , x2 2

3 1x, x 12 2

at1x2

28. For what value of k, will the function2x 16

if x 4f x x 4k i f x 4

be continuous at x = 4 ?

29. Determine the points of discontinuty, if any, of the following functions :

(a)2

2x 3

x x 1(b)

2

24x 3x 5x 2x 1

(b)2

2x x 1x 3x 1

(d)4x 16, x 2f(x)16, x 2

30. Show that the functions i n x cos, x 0f(x) x

2, x 0 is continuous at x = 0

31. Determine the value of 'a', so that the function f (x) defined by

acosx , x2x 2f(x)

5, x2

is continuous.

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218


ANSWERS

CHECK YOUR PROGRESS 20.11. (a) 17 (b) 7 (c) 0 (d) 2

(e)- 4 (f) 8

2. (a) 0 (b)32

(c)2

11(d)

qb

(e) 6

(f) 10 (g) 3 (h) 2

3. (a) 3 (b)72

(c) 4 (d)12

4. (a)12

(b)1

2 2 (c)1

2 6 (d) 2 (e) 1

5. (a) Does not exist (b) Does not exist

6. (a) 0 (b)14

(c) does not exist

7. (a)1, 2 (b)1 (c) 198. a 2

10. limit does not exist


1. (a) 2 (b)2

2e 1e 1

2. (a)1e

(b) e

3. (a) 2 (b)15

(c) 0 (d)ab

4. (a)12

(b)0 (c) 4 (d)23

5. (a)2

2ab

(b)2 (c)12

6. (a)1 (b)2

(c) 0

7. (a)53

(b)74

(c) -5

MATHEMATICS 219

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CHECK YOUR PROGRESS 20.31. (a) Continuous (b) Continuous

(c) Continuous (d) Continuous

5. (a) p =3 (b) a = 4 (c)14b9

CHECK YOUR PROGRESS 20.42. (a) Continuous

(b) Discontinuous at x = 2

(c) Discontinuous at x 3

(d) Discontinuous at x = 4

CHECK YOUR PROGRESS 20.51. (b) Continuous (c) Discontinuous

(d) Discontinuous (e)3k4

2 (a) Continuous (c) Continuous,(d) Discontinuous (e) Discontinuous

3 (a) Discontinuous (b) Continuous (c)53

4 (b) Continuous (c) k = 2

(d) Discontinuous

5. (a) Discontinuous (b) Discontinuous

6 (a) Continuous (b) Discontinuous

(c) Discontinuous (d) Continuous

7. (a) All real number except 1 and 4

(b) All real numbers except 2 and 3

(c) All real number except 6 and 1

(d) All real numbers except 4

TERMINAL EXERCISE

1. 5 2. 2 3. 4 4.13

5. 22x 6. 1 7.12

8.1

10

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220


9. 8 10.12

11. 1 12.a 12a

13. 1, 2 14. 2 ,2 15. 1 16. 1

17. 1 19. k 8 20.72

21. 1

22.92

23. 1 24.2

25.58

26. Discontinuous 27. Discontinuous28. k 829. (a) No (b) x 1(c) x 1, x 2 (d) x 231. 10

MATHEMATICS 221

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Differentiation

21

DIFFERENTIATION

The differential calculus was introduced sometime during 1665 or 1666, when Isaac Newtonfirst concieved the process we now know as differentiation (a mathematical process and it yieldsa result called derivative). Among the discoveries of Newton and Leibnitz are rules for findingderivatives of sums, products and quotients of composite functions together with many otherresults. In this lesson we define derivative of a function, give its geometrical and physicalinterpretations, discuss various laws of derivatives and introduce notion of second order derivativeof a function.

OBJECTIVESAfter studying this lesson, you will be able to :� define and interpret geometrically the derivative of a function y = f(x) at x = a;� prove that the derivative of a constant function f(x) = c, is zero;

� find the derivative of f n(x) x ,n Q from first principle and apply to find the derivativesof various functions;

� find the derivatives of the functions of the form cf(x), f x g x and polynomial functions;

� state and apply the results concerning derivatives of the product and quotient of two functions;� state and apply the chain rule for the derivative of a function;● find the derivative of algebraic functions (including rational functions); and● find second order derivative of a function.

EXPECTED BACKGROUND KNOWLEDGE● Binomial Theorem for any index● Functions and their graphs● Notion of limit of a function

21.1 DERIVATIVE OF A FUNCTIONConsider a function and a point say (5,25) on its graph. If x changes from 5 to 5.1, 5.01,

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Differentiation5.001..... etc., then correspondingly, y changes from 25 to 26.01, 25.1001, 25.010001,....Asmall change in x causes some small change in the value of y. We denote this change in the valueof x by a symbol x and the corresponding change caused in y by y and call these respectively

as an increment in x and increment in y, irrespective of sign of increment. The ratioxy of increment

is termed as incrementary ratio. Here, observing the following table for 2y x at (5,25), we

have for x 0.1 ,0.01,0.001, 0.0001,...... y 1.01,.1001, .010001, .00100001,.....x 5.1 5.01 5.001 5.0001

x .1 .01 .001 .0001

y 26.01 25.1001 25.010001 25.00100001

y 1.01 .1001 .010001 .00100001

yx

10.1 10.01 10.001 10.0001

We make the following observations from the above table :(i) y varies when x varies.(ii) y 0 when x 0 .

(iii) The ratio yx

tends to a number which is 10.

Hence, this example illustrates that y 0 when x 0 butyx

tends to a finite number, not

necessarily zero. The limit, x 0

ylim

xis equivalently represented by dy .

dxdydx

is called the derivative

of y with respect to x and is read as differential coefficient of y with respect to x.

That is, x 0

y dylim 10x dx

in the above example and note that while x and y are small

numbers (increments), the ratio yx

of these small numbers approaches a definite value 10.

In general, let us consider a functiony f(x) (i)

To find its derivative, consider x to be a small change in the value of x, so x x will be thenew value of x where f (x) is defined. There shall be a corresponding change in the value of y.Denoting this change by y ; y + y will be the resultant value of y, thus,

y y f(x x) ...(ii)

Subtracting (i) from (ii), we have,(y y) y f (x x) f(x)

or y f (x x) f(x) ...(iii)

To find the rate of change, we divide (iii) by x

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y f(x x) f(x)x x

...(iv)

Lastly, we consider the limit of the ratio yx

as x 0 .

Ifx 0 x 0

y f(x x) f(x)lim limx x

...(v)

is a finite quantity, then f(x) is called derivable and the limit is called derivative of f(x) with respect

to (w.r.t.) x and is denoted by the symbol f ' x or by d

dxof f(x)

i.e.d f(x)

dxor

dydx

(read as d

dxof y).

Thus,x 0 x 0


dy d f(x)dx dx

f '(x)

Remarks(1) The limiting process indicated by equation (v) is a mathematical operation. This

mathematical process is known as differentiation and it yields a result called a derivative.(2) A function whose derivative exists at a point is said to be derivable at that point.(3) It may be verified that if f (x) is derivabale at a point x = a, then, it must be continuous at

that point. However, the converse is not necessarily true.(4) The symbols x and h are also used in place of x i.e.

h 0

dy f(x h) f(x)limdx h

orx 0

dy f(x x) f(x)limdx x

(5) If y = f (x), then dydx

is also denoted by 1y or y ' .

21.2 VELOCITY AS LIMITLet a particle initially at rest at 0 moves along a strainght line OP., The distance s

O P Q

f (t) f (t + t)�

Fig. 21.1

covered by it in reaching P is a function of time t, We may write distanceOP s f(t) ...(i)

In the same way in reaching a point Q close to P covering PQi.e., s is a fraction of time t so that

OQ OP PQ

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Differentiation

s s

f(t t) ...(ii)

The average velocity of the particle in the interval t is given by

Change in distanceChange in time

(s+ s) s(t+ t) t , [From (i) and (ii)]

f (t t) f(t)t

( average rate at which distance is travelled in the interval t ).

Now we make t smaller to obtain average velocity in smaller interval near P. The limit of averagevelocity as t 0 is the instantaneous velocity of the particle at time t (at the point P).

Velocity at time t 0

f(t t) f(t)t limt

It is denoted by ds .dt

Thus, if f (t) gives the distance of a moving particle at time t, then the derivative of 'f' at 0t t

represents the instantaneous speed of the particle at the point P i.e. at time 0t t .

This is also referred to as the physical interpretation of a derivative of a function at a point.

Note : The derivative dydx

represents instantaneous rate of change of y w.r.t. x.

Example 21.1 The distance 's' meters travelled in time t seconds by a car is given bythe relation

2s 3tFind the velocity of car at time t =4 seconds.Solution : Here, 2f( t ) s 3t

2f ( t t) s s 3(t t)

Velocity of car at any timet 0

f(t t) f(t)t limt

2 2

t 0

3(t t) 3tlimt

2 2 2

t 0

3(t 2t. t t ) 3tlimt

t 0lim (6t 3 t)

= 6t

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Differentiation Velocity of the car at t = 4 sec

= (6 × 4) m/sec

= 24 m/sec.

CHECK YOUR PROGRESS 21.11. Find the velocity of particles moving along a straight line for the given time-distance relations

at the indicated values of time t :

(a)1s 2 3t;t .3

(b) s 8t 7;t 4.

(c) 2 3s t 3t;t .2

(d) 2 5s 7t 4t 1;t .2

2. The distance s metres travelled in t seconds by a particle moving in a straight line is given

by 4 2s t 18t . Find its speed at t = 10 seconds.

3. A particle is moving along a horizontal line. Its distance s meters from a fixed point O at tseconds is given by 2 3s 10 t t . Determine its instantaneous speed at the end of 3seconds.

21.3 GEOMETRICAL INTERPRETATION OF dy/dxLet y = f (x) be a continuous function of x, draw its graph and denote it by APQB.

Fig. 21.2

Let P (x,y) be any point on the graph of y = f(x) or curve represented by y = f(x). LetQ(x x,y y) be another point on the same curve in the neighbourhood of point P..

Draw PM and QN perpendiculars to x-axis and PR parallel to x-axis such that PR meets QN atR.Join QP and produce the secant line to any point S. Secant line QPS makes angle say withthe positive direction of x-axis. Draw PT tangent to the curve at the point P, making angle withthe x-axis.

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Differentiation

Now, In QPR, QPR

QR QN RN QN PM (y y) y ytanPR MN ON OM (x x) x x (i)

Now, let the point Q move along the curve towards P so that Q approaches nearer and nearerthe point P.Thus, when Q P , x 0 , y 0 , 0,(tan tan ) and consequently, the secant QPStends to coincide with the tangent PT.

From (i).ytanx

In the limiting case, x 0 x 0y 0

ylim tan limx

ordytandx

...(ii)

Thus the derivative dydx

of the function y = f (x) at any point P (x,y) on the curve represents the

slope or gradient of the tangent at the point P.

This is called the geometrical interpretation of dydx

.

It should be noted that dydx

has different values at different points of the curve.

Therefore, in order to find the gradient of the curve at a particular point, find dydx

from the

equation of the curve y = f (x) and substitute the coordinates of the point in dydx

.

Corollary 1

If tangent to the curve at P is parallel to x-axis, then 0 or180°, i.e.,dy tan0dx

or tan

180° i.e., dy 0dx

.

That is tangent to the curve represented by y = f (x) at P is parallel to x-axis.

Corollary 2

If tangent to the curve at P is perpendicular to x-axis, 90 ordy tan90dx

.

That is, the tangent to the curve represented by y = f (x) at P is parallel to y-axis.

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21.4 DERIVATIVE OF CONSTANT FUNCTIONStatement : The derivative of a constant is zero.

Proof : Let y = c be a constant function. Then y = c can be written as0y cx 0[ x 1]∵ ..(i)

Let x be a small increment in x. Corresponding to this increment, let y be the increment in thevalue of y so that

0y y c(x x) ..(ii)Subtracting (i) from (ii),

0 0(y y) y c(x x) cx , 0x 1∵or y c c or y 0

Dividing by x ,y 0x x

ory 0x

Taking limit as x 0, we have

x 0

ylim 0x

ordy 0dx

ordc 0dx

[ y = c from (i)]

This proves that rate of change of constant quantity is zero. Therefore, derivative of a constantquantity is zero.

21.5 DERIVATIVE OF A FUNCTION FROM FIRST PRINCIPLERecalling the definition of derivative of a function at a point, we have the following working rulefor finding the derivative of a function from first principle:

Step I. Write down the given function in the form of y = f (x) ....(i)Step II. Let dx be an increment in x, y be the corresponding increment in y so that

y y f(x x) ....(ii)

Step III. Subtracting (i) from (ii), we gety f (x x) f(x) ..(iii)

Step IV. Dividing the result obtained in step (iii) by x , we get,y f (x x) f(x)x x

Step V. Proceeding to limit as x 0 .

x 0 x 0


Note : The method of finding derivative of function from first principle is also called deltaor ab-ininitio method.

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Differentiation

Next, we find derivatives of some standard and simple functions by first principle.

21.6 DERIVATIVES OF THE FUNCTIONS FROMTHE FIRST PRINCIPLE

nLety x .....(i)For a small increment x in x, let the corresponding increment in y be y .

Then ny y (x x) . ...(ii)Subtracing (i) from (ii) we have,

n n(y y) y (x x ) x

nn nx

y x 1 xx

nn xx 1 1

x

Sincex

x<1, as x is a small quantity compared to x, we can expand

nx1

xby Binomial

theorem for any index.

Expandingnx

1x

by Binomial theorem, we have

2 3n x n(n 1) x n(n 1)(n 2) x

y x 1 n ... 1x 2! x 3! x

2n

2 3n n 1 n n 1 n 2 xn x

x x ....x 2 3!x x

Dividing by x , we have2

n2 3

y n n(n 1) x n(n 1)(n 2) ( x)x ...

x x 2! 3!x x

Proceeding to limit when x 0 , 2( x) and higher powers of x will also tend to zero.

2n

2 3x 0 x 0

y n n(n 1) x n(n 1)(n 2) ( x)lim limx ...

x x 2! 3!x x

orn

x 0

y dy nlt x 0 0 ...

x dx x

or n n 1dy nx nxdx x

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or n n 1d (x ) nx ,dx

ny x∵

This is known as Newton's Power Formula or Power Rule

Note : We can apply the above formula to find derivative of functions like 2 3x,x , x ,...

i.e. when n = 1,2,3,...

e.g. 1 1 1 0d dx x 1x 1x 1.1 1dx dx

2 2 1d x 2x 2xdx

3 3 1 2d x 3x 3x ,dx

and so on.

Example 21.2 Find the derivative of each of the following :

(i) 10x (ii) 50x (iii) 91xSolution :

(i) 10 10 1 9d x 10x 10xdx

(ii) 50 50 1 49d x 50x 50xdx

(iii) 91 91 1 90d x 91x 91xdx

Remark : Newton's power formula is also applicable when n is a rational number.

i.e. say when n=12

. For example,

Let12y x x

Then1 112 2

1 / 2dy 1 1 1 1x xdx 2 2 2 x

ord 1x

dx 2 x

We shall now find the derivatives of some simple functions from definition or first principles.

Example 21.3 Find the derivative of 2x from the first principles.

Solution : Let 2y x (i)

For a small increment x in x let the corresponding increment in y be y .2y y (x x) (ii)

Subtracting (i) from (ii), we have

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Differentiation

2 2(y y) y (x x) x

or 2 2 2y x 2x( x) ( x) x

or 2y 2x( x) ( x)Divide by x , we have

y 2x xx

Proceeding to limit when x 0 , we have

x 0 x 0

ylim lim (2x x)x

orx 0

dy 2x lim ( x)dx

= 2x + 0= 2x

ordy 2xdx

or 2d x 2xdx

Example 21.4 Find the derivative of 1 , x 0x

by delta method.

Solution : Let1yx

(i)

For a small increment x in x, let the corresponding increment in y be y .

1y yx x

(ii)

Subtracting (i) from (ii), we have,1 1(y y) y

x x x

orx (x x)y(x x)x

xx(x x) (iii)

Dividing (iii) by x , we have

y 1x x(x x)

Proceeding to limit as x 0 , we have

x 0 x 0

y 1lim limx x(x x)

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Differentiation

ordy 1dx x(x 0) or 2

d 1 1dx x x

Example 21.5 Find the derivative of x by ab-initio method.

Solution : Let y x ..(i)

For a small increment x in x, let y be the corresponding increment in y..

y y x x ...(ii)

Subtracting (i) from (ii), we havey y y x x x ...(iii)

or y x x x

Rationalising the numerator of the right hand side of (iii), we have

x x xy x x x

x x x

(x x) x

x x xor

xyx x x

Dividing by x , we have

y 1x x x x

Proceeding to limit as x 0 , we have

x 0 x 0

y 1lim limx x x

ordy 1dx x x or

d 1xdx 2 x

Example 21.6 If f (x) is a differentiable function and c is a constant, find the derivative of(x) cf(x)

Solution : We have to find derivative of function(x) cf(x) ...(i)

For a small increment x in x, let the values of the functions (x)be (x x) and that of f(x)be f(x x)

(x x) cf(x x) ...(ii)

Subtracting (i) from (ii), we have(x x) (x) c f (x x) f(x)

Dividing by x , we have

(x x) (x) f (x x) f(x)c

x x

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DifferentiationProceeding to limit as x 0 , we have

x 0 x 0

(x x) (x) f (x x) f(x)lim lim c

x x

orx 0

f(x x) f(x)'(x) c lim

x

or '(x) cf '(x)

Thus,d dfcf x c

dx dx


1. Find the derivative of each of the following functions by delta method :

(a) 10x (b) 2x 3 (c) 23x

(d) 2x 5x (e) 37x

2. Find the derivative of each of the following functions using ab-initio method:

(a)1 , x 0x

(b)1 ,x 0

ax(c) 1x ,x 0

x

(d)1 b,x

ax b a(e)

ax b d,xcx d c

(f)x 2 5,x

3x 5 3

3. Find the derivative of each of the following functions from first principles :

(a)1 , x 0x (b)

1 b, xaax b (c)

1x , x 0x

(d)1 x , x 11 x

4. Find the derivative of each of the following functions by using delta method :(a) f(x) 3 x . Also find f '(2). (b) 2f(r) r . Also find f '(2).

(c) 34f(r) r3

. Also find f '(3).

21.7 ALGEBRA OF DERIVATIVESMany functions arise as combinations of other functions. The combination could be sum,difference, product or quotient of functions. We also come across situations where a givenfunction can be expressed as a function of a function.In order to make derivative as an effective tool in such cases, we need to establish rules forfinding derivatives of sum, difference, product, quotient and function of a function. These, inturn, will enable one to find derivatives of polynomials and algebraic (including rational) functions.

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21.8 DERIVATIVES OF SUM AND DIFFERENCE OF FUNCTIONS

If f (x) and g (x) are both derivable functions and h (x) = f (x) + g ( x), then what is h' (x) ?Here h (x) = f (x) + g (x)

Let x be the increment in x and y be the correponding increment in y..h(x x) f(x x) g(x x)

Hencex 0

f(x x) g(x x) f(x) g(x)h'(x) lim

x

x 0

f(x x) f(x) g(x x) g(x)lim

x

x 0

f(x x) f(x) g(x x) g(x)lim

x x

x 0 x 0

f x x f x g x x g xlim lim

x x

or h ' x f ' x g ' x

Thus we see that the derivative of sum of two functions is sum of their derivatives.

This is called the SUM RULE.

e.g. 2 3y x x

Then 2 3d dy ' x xdx dx

22x 3x

Thus 2y ' 2x 3x

This sum rule can easily give us the difference rule as well, because

if h x f x g x

then h x f x g x

h ' x f ' x g ' x

f ' x g ' x

i.e. the derivative of difference of two functions is the difference of their derivatives.

This is called DIFFERENCE RULE.Thus we have

Sum rule :d d df x g x f x g x

dx dx dx

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Differentiation

Difference rule :d d df x g x f x g x

dx dx dx

Example 21.7 Find the derivative of each of the following functions :

(i) 2 3y 10t 20t

(ii) 3 2y 2x 3x

(iii)3

21 1y x

xx, x 0

Solution :

(i) We have, 2 3y 10t 20t

2dy 10 2t 20 3tdt

220t 60t(ii) 3 2y 2x 3x

2dy 6x 6xdx

(iii) 32

1 1y xxx

x 0

3 2 1x x x

2 3 2dy 3x ( 2)x ( 1)xdx

23 2

2 13xx x

Example 21.8 Evaluate the derivative of each of the following functions at the indicated

value (s) :

(i) 2s 4.9t 2.4, t 1,t 5 (ii) 3 2y x 3x 4x 5, x 1

Solution : (i) We have 2s 4.9t 2.4

ds 4.9(2t)dt = 9.8 t

t 1

ds 9.8 1 9.8dt

t

ds 9.8 5 49dt

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Differentiation

(ii) We have 3 2y x 3x 4x 5

3 2 2dy d x 3x 4x 5 3x 6x 4dx dx

2

x 1

dy 3 1 6 1 4dx

= 13

CHECK YOUR PROGRESS 21.31. Find y' when :

(a) y =12 (b) y = 12x (c) y = 12 x + 12

2. Find the derivatives of each of the following functions :

(a) 9f(x) 20x 5x (b) 4 2f(x) 50x 20x 4

(c) 3 2f(x) 4x 9 6x (d) 95f(x) x 3x9

(e) 3 2 2f(x) x 3x 3x5

(f)8 6 4x x xf(x) 2

8 6 4

(g)2 43 5

22 3f(x) x x5 x

(h)1f(x) xx

3. (a) If f(x) 16x 2 , find f '(0),f'(3),f'(8)

(b) If3 2x xf(x) x 16

3 2, find f ' 1 , f ' 0 , f ' 1

(c) If4

7x 3f(x) x 2x 54 7

, find f ' 2

(d) Given that 34V r3

, find dVdr

and hencer 2

dVdr

21.9 DERIVATIVE OF PRODUCT OF FUNCTIONSYou are all familiar with the four fundamental operations of Arithmetic : addition, subtraction,multiplication and division. Having dealt with the sum and the difference rules, we now considerthe derivative of product of two functions.Consider 2 2y (x 1)

This is same as 2 2y (x 1)(x 1)

So we need now to derive the way to find the derivative in such situation.We write 2 2y (x 1)(x 1)

Let x be the increment in x and y the correrponding increment in y. Then

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Differentiation2 2y y [(x x) 1][(x x) 1)]

2 2 2 2y [(x x) 1][(x x) 1)] (x 1)(x 1)

2 2 2[(x x ) 1][(x x) x )] 2 2 2 2(x 1)[(x x) 1] (x 1)(x 1)

2 2 2 2 2 2[(x x ) 1][(x x) x ] (x 1)[x x) 1 (x 1)]

2 2 2 2 2 2[(x x) 1][(x x) x ] (x 1)[(x x) x ]

2 22 22 2x x x x x xy x x 1 x 1

x x x

2 22 22x x x 2x x x

x x 1 x 1x x

2 2[(x x) 1](2x x) (x 1)(2x x)

2 2x 0 x 0 x 0

ylim lim [(x x) 1] [2x x] lim (x 1)(2x x)x

or 2 2dy x 1 2x x 1 2xdx

24x x 1

Let us analyse : 2 2

derivative derivative2 2of x 1 o f x 1

dy x 1 2x x 1 2xdx

Consider 3 2y x x

Is 3 2 2dy x 2x x 3xdx

?

Let us check 3 2 2x 2x x 3x

4 42x 3x45x

We have 3 2y x x

5x

4dy 5xdx

In general, if f (x) and g (x) are two functions of x then the derivative of their product is definedby

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Differentiation

d f x g x f x g ' x g x f ' xdx

d dIst function Secondfunction Secondfunction Istfunction

dx dxwhich is read as derivative of product of two functions is equal to

= [Ist function] [Derivative of Second function] +

[Second function] [Derivative of Ist function]

This is called the PRODUCT RULE.

Example 21.9 Find dy ,dx

if 6 2y 5x 7x 4x

Method I. Here y is a product of two functions.

6 2 2 6dy d d5x 7x 4x 7x 4x 5xdx dx dx

6 2 55x 14x 4 7x 4x 30x

7 6 7 670x 20x 210x 120x7 6280x 140x

Method II 6 2y 5x 7x 4x8 735x 20x

7 6 7 6dy 35 8x 20 7x 280x 140xdx

which is the same as in Method I.This rule can be extended to find the derivative of two or more than two functions.

Remark : If f (x), g (x) and h(x) are three given functions of x, then

d d d df x g x h x f x g x h x g x h x f x h x f x g xdx dx dx dx

Example 21.10 Find the derivative of f x g x h x if

f x x,g x x 3 , and 2h x x x

Solution : Let y = 2x x 3 x x

To find the derivative of y, we can combine any two functions, given on the R.H.S. and apply theproduct rule or use result mentioned in the above remark.In other words, we can write

2y x x 3 x x

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Differentiation

Let u x f x g x

x x 3

2x 3x

Also 2h x x x

y u x h x

Hence 2 2 2dy d dx x 3 x x x x x 3xdx dx dx

2x x 3 2x 1 x x 2x 3

2 2x x 3 2x 1 x x x 3 x x x

[f x g x ] h ' x [g x h(x)]f'(x) [h x f(x)].g'(x)

Henced d[f x g x h x ] [f x g(x)] h(x)

dx dx

d d[g x h(x)] f(x) h(x)f(x) g(x)dx dx

Alternatively, we can directly find the derivative of product of the given three functions.

2 2 2dy d d d[x x 3 ] (x x) [ x 3 (x x)] (x) [(x x) x] (x 3)dx dx dx dx

2 2x x 3 (2x 1) x 3 (x x) 1 (x x) x 1

3 24x 6x 6x

CHECK YOUR PROGRESS 21.41. Find the derivative of each of the following functions by product rule :

(a) f x (3x 1)(2x 7) (b) f(x) (x 1)( 3x 2)

(c) f(x) (x 1)( 2x 9) (d) y (x 1)(x 2)

(e) 2 2y x (2x 3x 8) (f) 2y (2x 3)(5x 7x 1)

(g) 2 3u(x) (x 4x 5)(x 2)

2. Find the derivative of each of the functions given below :

(a) 2f(r) r(1 r)( r r) (b) f(x) (x 1)(x 2)(x 3)

(c) 2 3 2 4f(x) (x 2)(x 3x 4)(x 1)

(d) 2 2f(x) (3x 7)(5x 1) 3x 9x 8

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Differentiation

21.10 QUOTIENT RULEYou have learnt sum Rule, Difference Rule and Product Rule to find derivative of a functionexpressed respectively as either the sum or difference or product of two functions. Let us nowtake a step further and learn the "Quotient Rule for finding derivative of a function which is thequotient of two functions.

Let1g(x)

r(x) , [r(x) 0]

Let us find the derivative of g (x) by first principles1g(x)

r(x)

x 0

1 1r(x x) r(x)g'(x) lim

x

x 0

r(x) r(x x)lim(x)r(x)r(x x)

x 0 x 0

r(x) r(x x) 1lim limx r(x).r(x x)

2 21 r'(x)r'(x)

[r(x)] [r(x)]

Consider any two functions f (x) and g (x) such that f(x)(x)g(x)

, g x 0

We can write 1(x) f xg(x)

1 d 1(x) f ' x f(x)

g(x) dx g(x)

2f'(x) g'(x)

f(x)g(x) [g(x)]

2g(x)f'(x) f(x)g'(x)

[g(x)]

(Denominator)(Derivative of Numerator) (Numerator)(Derivative of Denominator)2(Denominator)

Hence 2d f(x) f '(x)g(x) f(x)g'(x)

dx g(x) [g(x)]

This is called the quotient Rule.

MATHEMATICS

Notes

MODULE - VCalculus

240

Differentiation

Example 21.11 Find f '(x)if4x 3f(x)2x 1

,1x2

Solution :

2

d d(2x 1) (4x 3) (4x 3) (2x 1)dx dxf '(x)

(2x 1)

2(2x 1).4 (4x 3).2

(2x 1)

210

(2x 1)Let us consider the following example:

Let1f(x)

2x 1,

1x2

2

d d(2x 1) (1) 1 (2x 1)d 1 dx dxdx 2x 1 (2x 1)

2(2x 1) 0 2

(2x 1)d (1) 0

dx∵

i.e. 2d 1 2

dx 2x 1 (2x 1)

CHECK YOUR PROGRESS 21.51. Find the derivative of each of the following :

(a)2y

5x 7,

7x5

(b) 23x 2y

x x 1(c)

2

2x 1yx 1

(d)4

2xf(x)

x 3(e)

5

7x 2xf(x)

x(f) 2

xf(x)x x 1

(g) 3x

f(x)x 4

2. Find f '(x) if

(a)2x(x 3)f(x)

x 2, x 2

(b) (x 1) x 2f(x)

(x 3)(x 4), x 3, x 4

MATHEMATICS 241

Notes

MODULE - VCalculus

Differentiation

21.11 CHAIN RULEEarlier, we have come across functions of the type 4 2x 8x 1 . This function can not beexpressed as a sum, difference, product or a quotient of two functions. Therefore, the techniquesdeveloped so far do not help us find the derivative of such a function. Thus, we need to developa rule to find the derivative of such a function.

Let us write : 4 2y x 8x 1 or y t where 4 2t x 8x 1

That is, y is a function of t and t is a function of x. Thus y is a function of a function. We proceedto find the derivative of a function of a function.Let t be the increment in t and y , the corresponding increment in y..Then y 0 as t 0

t 0

dy ylimdt t

(i)

Similarly t is a function of x.t 0 as x 0

x 0

dt tlimdx x

(ii)

Here y is a function of t and t is a function of x. Therefore y 0 as x 0


x 0 t 0 x 0

dy y y tlim lim lim

dx x t x

dy dtdt dx

Thusdy dy dtdx dt dx

This is called the Chain Rule.

Example 21.12 If 4 2y x 8x 1 , find dydx

Solution : We are given that4 2y x 8x 1

which we may write asy t , where 4 2t x 8x 1 (i)

dy 1dt 2 t and 3dt 4x 16x

dx

Here 3dy dy dt 1 (4x 16x)dx dt dx 2 t

MATHEMATICS

Notes

MODULE - VCalculus

242

Differentiation

3 3

4 2 4 2

4x 16x 2x 8x

2 x 8x 1 x 8x 1(Using (i))

Example 21.13 Find the derivative of 3 2y x 5x 7

Solution : We have3 2y x 5x 7

3 2 1/3 2d dx 5x 7 t where t x 5x 7dx dx

(i)

1/3d dttdt dx

(Using chain rule)

2 / 31 t (2x 5)3

3 2 2 2 / 3d 1x 5x 7 [x 5x 7] (2x 5)dx 3

[Using (i)]

Example 21.14 Find the derivative of the function 2 7

5y(x 3)

Solution : 2 7dy d 5(x 3)dx dx

2 8 2d5[( 7)(x 3) ] (x 3)dx

(Using chain Rule)

2 835(x 3) (2x)

2 870x

(x 3)

Example 21.15 Find 4 3dy 1 2where y v and v x 5dx 4 3

Solution : We have 4 31 2y v and v x 54 3

3dy 1 (4v )dv 4

=3

3 32v x 5

3...(i)

and 2dv 2 (3x )dx 3

22x ..(ii)

Thusdy dy dvdx dv dx

MATHEMATICS 243

Notes

MODULE - VCalculus

Differentiation

33 22

x 5 (2x )3

[Using (i) and (ii)]

Remark

We have seen in the previous examples that by using various rules of derivatives we can findderivatives of algebraic functions.

CHECK YOUR PROGRESS 21.61. Find the derivative of each of the following functions :

(a) 7f(x) (5x 3) (b) 2 35f(x) (3x 15)

(c) 2 17f(x) (1 x ) (d)53 x

f(x)7

(e)2

1yx 3x 1

(f) 2 53y (x 1)

(g)2

1y7 3x

(h)5

6 41 1 1y x x

6 2 16

(i) 2 4y (2x 5x 3) (j) 2y x x 8

2. Finddydx

if

(a) 23 v 4xy , v2 v 1 x

(b) 2 xy at , t2a

21.12 DERIVATIVES OF A FUNCTION OF SECOND ORDER

Second Order Derivative : Given y is a function of x, say f (x). If the derivative dydx

is a

derivable function of x, then the derivative of dydx

is known as the second derivative of y = f (x)

with respect to x and is denoted by 2

2d ydx

. Other symbols used for the second derivative of y are

2D , f", y", 2y etc.

Remark

Thus the value of f " at x is given by

h 0

f ' x h f ' hf " x lim

h

MATHEMATICS

Notes

MODULE - VCalculus

244

Differentiation

The derivatives of third, fourth, ....orders can be similarly defined.

Thus the second derivative, or second order derivative of y with respect to x is2

2d dy d y

dx dx dx

Example 21.16 Find the second order derivative of

(i) 2x (ii) 3x +1 (iii) 2(x 1)(x 1) (iv)x 1x 1

Solution : (i) Let 2y x , then dy 2xdx

and2

2d y d d(x)(2x) 2

dx dxdx = 2.1 = 2

2

2d y 2dx

(ii) Let 3y x 1 , then

2dy 3xdx

(by sum rule and derivative of a constant is zero)

and2

22

d y d (3x ) 3.2x 6xdxdx

2

2d y 6xdx

(iii) Let 2y (x 1)(x 1), then

2 2dy d d(x 1) (x 1) (x 1), (x 1)dx dx dx

2(x 1) 1 (x 1) 2x or 2 2 2dy x 1 2x 2x 3x 2x 1dx

and2

22

d y d 3x 2x 1 6x 2dxdx

2

2d y 6x 2dx

(iv) Letx 1yx 1

, then

2 2dy (x 1) 1 (x 1).1 2dx (x 1) (x 1)

MATHEMATICS 245

Notes

MODULE - VCalculus

Differentiation

and2

2 2 3 3d y d 2 1 4

2. 2.dxdx x 1 x 1 x 1

2

2 3d y 4dx x 1

CHECK YOUR PROGRESS 21.7Find the derivatives of second order for each of the following functions :

(a) 3x (b) 4 3 2x 3x 9x 10x 1

(c)2x 1

x 1(d) 2x 1

� The derivative of a function f (x) with respect to x is defined as

x 0

f x x f xf ' x lt , x 0

x

� The derivative of a constant is zero i.e., dc 0dx

, where c a is constant.

� Newton's Power Formula

n n 1d x nxdx

� Geometrically, the derivative dydx

of the function y f x at point P x, y is the slope

or gradient of the tangent on the curve represented by y f x at the point P..� The derivative of y with respect to x is the instantaneous rate of change of y with respect

to x.� If f (x) is a derivable function and c is a constant, then

d cf x c f ' xdx

, where f ' x denotes the derivative of f x .

� 'Sum or difference rule' of functions :d d df x g x f x g x

dx dx dxDerivative of the sum or difference of two functions is equal to the sum or diference oftheir derivatives respectively.

LET US SUM UP

MATHEMATICS

Notes

MODULE - VCalculus

246

Differentiation� Product rule :

d d df x g x f x g x g x f xdx dx dx

d dIst function IInd function IIndfunction Ist function

dx dx

� Quotient rule : Iff x

x , g x 0, theng x

2g x f ' x f x g ' x

' xg x

2

d dDenominator Numerator Numerator Denominatordx dx

Denominator

� Chain Rule :d df g(x) f '[g(x)] [g(x)]

dx dx

=derivative of f(x) w.r.t g (x)×derivative of g (x) w.r.t.x

� The derivative of second order of y w.r.t. to x is 2

2d dy d y

dx dx dx


TERMINAL EXERCISE

1. The distance s meters travelled in time t seconds by a car is given by the relation 2s t .Caclulate.(a) the rate of change of distance with respect to time t.(b) the speed of car at time t = 3 seconds.

2. Given 2f( t ) 3 4t . Use delta method to find 1

f ' ( t ) , f '3 .


MATHEMATICS 247

Notes

MODULE - VCalculus

Differentiation

3. Find the derivative of 4f(x) x from the first principles. Hence find

1f '(0),f'

2

4. Find the derivative of the function 2x 1 from the first principles.

5. Find the derivatives of each of the following functions by the first principles :

(a) ax + b, where a and b are constants (b) 22x 5

(c) 3 2x 3x 5 (d) 2(x 1)6. Find the derivative of each of the following functions :

(a) 4 2f(x) px qx 7x 11 (b) 3 2f(x) x 3x 5x 8

(c)1f(x) xx

(d)2x af(x) ,a 2

a 27. Find the derivative of each of the functions given below by two ways, first by product

rule, and then by expanding the product. Verify that the two answers are the same.

(a) 1y x 1x

(b)32 1y x 2 5x

x

8. Find the derivative of the following functions :

(a) 2xf(x)

x 1(b) 2 3

3 10f(x)(x 1) x

(c) 41f(x)

(1 x )(d)

(x 1)(x 2)f(x)x

(e)23x 4x 5f(x)

x(f) x 4f(x)

2 x

(g)3

2(x 1)(x 2)f(x)

x

9. Use chain rule, to find the derivative of each of the functions given below :

(a)21x

x(b)

1 x1 x (c) 2 23 x (x 3)

10. Find the derivatives of second order for each of the following :

(a) x 1 (b) x x 1

MATHEMATICS

Notes

MODULE - VCalculus

248

Differentiation

(a) x 1 (b) x x 1

ANSWERS

CHECK YOUR PROGRESS 21.11. (a) 3 (b) 8 (c) 6 (d) 31

2. 3640 m/s 3. 21 m/s


1. (a) 10 (b) 2 (c) 6x (d) 2x+5 (e) 221x

2. (a) 21x

(b) 21

ax(c) 2

11x

(d) 2a

(ax b) (e) 2ad bc

(cx d)

(f) 21

(3x 5)

3. (a)1

2x x (b)a

2(ax b)( ax b) (c)1 1

1x2 x

(d) 22

(1 x)

4. (a)3 3;

2 x 2 2 (b)2 r ; 4 (c) 22 r ;36

CHECK YOUR PROGRESS 21.3CHECK YOUR PROGRERSS 21.3

1. (a) 0 (b) 12 (c) 12

2. (a) 8180x 5 (b) 3200x 40x (c) 212x 12x

(d) 85x 3 (e) 23x 6x 3 (f) 7 5 3x x x

(g)1 9

33 54 4x x 6x15 5

(h) 32

1 12 x

2x3. (a) 16, 16, 16 (b) 3,1 ,1 (c) 186

(d) 24 r ,16

CHECK YOUR PROGRESS 21.41. (a)12x 19 (b) 6x 5 (c) 4x 11

(d)2x 3 (e) 3 28x 9x 16x (f) 230x 2x 19

MATHEMATICS 249

Notes

MODULE - VCalculus

Differentiation

(g) 4 3 25x 16x 15x 4x 8

2. (a) 3 24 r 3( 1)r 2r (b) 23x 12x 11

(c) 8 7 6 5 4 3 29x 28x 14x 12x 5x 44x 6x 4x

(d) 2 2 2(5x 1)(3x 9x 8).6x 5(3x 7)(3x 9x 8) 2(3x 7)(5x 1)(6x 9)


1. (a) 210

(5x 7) (b)2

2 23x 4x 1

(x x 1)(c) 2 2

4x(x 1)

(d)5 3

2 22x 12x(x 3)

(e)4

72x 12

x(f)

2

2 21 x

(x x 1)(g)

3

3 24 5x

2 x(x 4)

2. (a)3 2

22x 6x 6

(x 2)(b)

2

2 24x 20x 22

(x 3) (x 4)

CHECK YOUR PROGRESS 21.61. (a) 635(5x 6) (b) 2 34210x(3x 15)

(c) 2 1634x(1 x ) (d) 45(3 x)7

(e) 2 2(2x 3)(x 3x 1) (f)2

2 310x x 13

(g) 2 3 / 23x(7 3x ) (h)46 4

5 3 x x 15(x 2x )

6 2 16

(i) 2 54(4x 5)(2x 5x 3) (j) 2x1

x 8

2. (a)2

2 25(1 x )

(1 2x x )(b)

x2a


1. (a) 6x (b) 212x 18x 18 (c) 34

(x 1) (d) 2 3 / 21

(1 x )

TERMINAL EXERCISE1. (a) 2 t (b) 6 seconds

MATHEMATICS

Notes

MODULE - VCalculus

250

Differentiation

288t,3

3.10,

24.

12x 1

5. (a) a (b) 4x (c) 23x 6x (d)2(x 1)

6. (a) 34px 2qx 7 (b) 23x 6x 5

(c)2

11x

(d)2x

a 2

7. (a)1

2 x (b) 25 13 x x x2 2 x

8. (a)2

2 2(x 1)

(x 1)(b) 3 4

6 30(x 1) x

(c)3

4 24x

(1 x )(d) 3 / 2

3 1 1x2 2 x x

(e)2

53x

(f)1 1

4 x x x

(g) 22 3

1 43x 2x x

9. (a)2

11x

(b) 32

1

1 x (1 x )

(c)3

24 2 3

4x 6x

3(x 3x )

10. (a) 32

1

4(x 1)(b)

2

12

2 x x

4(x 1)

MATHEMATICS 251

Notes

MODULE - VCalculus

Differentiation of Trigonometric Functions

22

DIFFERENTIATION OF TRIGONOMETRICFUNCTIONS

Trigonometry is the branch of Mathematics that has made itself indispensable for other branchesof higher Mathematics may it be calculus, vectors, three dimensional geometry, functions-harmonicand simple and otherwise just cannot be processed without encountering trigonometric functions.Further within the specific limit, trigonometric functions give us the inverses as well.The question now arises : Are all the rules of finding the derivatives studied by us so far appliacableto trigonometric functions ?This is what we propose to explore in this lesson and in the process, develop the formulae orresults for finding the derivatives of trigonometric functions and their inverses . In all discussionsinvolving the trignometric functions and their inverses, radian measure is used, unless otherwisespecifically mentioned.

. OBJECTIVES

After studying this lesson, you will be able to :

� find the derivative of trigonometric functions from first principle;� find the derivative of inverse trigonometric functions from first principle;� apply product, quotient and chain rule in finding derivatives of trigonometric and inverse

trigonometric functions; and� find second order derivative of a function.

EXPECTED BACKGROUND KNOWLEDGE� Knowledge of trigonometric ratios as functions of angles.� Standard limits of trigonometric functions namely.

(i)x 0limsinx 0 (ii)

x 0

sinxlim 1x

(iii) x 0limcosx 1 (iv)

x 0

tanxlim 1x

� Definition of derivative, and rules of finding derivatives of function.

MATHEMATICS

Notes

MODULE - VCalculus

252


22.1 DERIVATIVE OF TRIGONOMETRIC FUNCTIONSFROM FIRST PRINCIPLE

(i) Let y = sin xFor a small increment x in x, let the corresponding increment in y be y .

y y sin(x x)

and y sin(x x) sinx

x x2cos x sin

2 2C D C D

sinC sinD 2cos sin2 2

xsiny x 22cos xx 2 x

x 0 x 0 x 0

xsiny x 2lim lim cos x lim xx 22

cosx.1 x 0

sin x2lim 1x2

∵

Thus,dy cosxdx

i.e.,d (sinx) cosx

dx(ii) Let y = cos xFor a small increment x in x, let the corresponding increment in y be y .

y y cos(x x)

and y cos(x x) cosx

x x2sin x sin

2 2

xsiny x 22sin xx 2 x

x 0 x 0 x 0

xsiny dx 2lim lim sin x lim xx 22

sinx 1

Thus,dy sinxdx

MATHEMATICS 253

Notes

MODULE - VCalculus


i.e.,d cosx sinx

dx(iii) Let y = tan xFor a small increament x in x, let the corresponding increament in y be y .

y y tan(x x)

and y tan(x x) tanx

sin(x x) sinxcos(x x) cosx

sin(x x) cosx sinx.cos(x x)cos(x x)cosx

sin[(x x) x ]cos(x x)cosx

sin xcos(x x) cosx

y sin x 1x x cos(x x)cosx

or x 0 x 0 x 0

y sin x 1lim lim limx x cos(x x)cosx

211

cos x x 0

sin xlim 1

x∵

2sec x

Thus, 2dy sec xdx

i.e. 2d (tanx) sec xdx

(iv) Let y = sec xFor a small increament x in x, let the corresponding increament in y be y .

y y sec(x x)and y sec(x x) secx

=1 1

cos(x x) cosx

cosx cos(x x)cos(x x)cosx

x x2sin x sin2 2

cos(x x)cosx

MATHEMATICS

Notes

MODULE - VCalculus

254


x 0 x 0

x xsin x siny 2 2lim lim xx cos(x x)cosx2

x 0 x 0 x 0

x xsin x siny 2 2lim lim lim xx cos(x x)cosx2

2sinx 1

cos x

sinx 1 tanx.secxcosx cosx

Thus,dy secx.tanxdx

i.e.d (secx) secx tanx

dxSimilarly, we can show that

2d (cotx) cosec xdx

andd (cosec x ) cosec x cot x

dx

Example 22.1 Find the derivative of 2cotx from first principle.

Solution : 2y cotx

For a small increament x in x, let the corresponding increament in y be y .2y y cot(x x)

and 2 2y cot(x x) cotx

2 2

2 2cos(x x) cosxsin(x x) sinx

2 2 2 2

2 2cos(x x) sinx cosx sin(x x)

sin(x x) sinx

2 2

2 2sin[x (x x) ]sin(x x) sinx

2

2 2sin[ 2x x ( x) ]sin(x x) sinx

MATHEMATICS 255

Notes

MODULE - VCalculus


2 2sin[(2x x) x]

sin(x x) sinx

2 2y sin[(2x x) x]x xsin(x x) sinx

and 2 2x 0 x 0 x 0

y sin[(2x x) x] 2x xlim lim limx x(2x x) sin(x x) sinx

ordy 1dx 2 2

2xsinx .sinx x 0

sin[(2x x) x]lim 1

x(2x x)∵

2 2 2 22x 2x

(sinx ) sin x2 22x.cosec x

Hence 2 2 2d (cotx ) 2x cosec xdx

Example 22.2 Find the derivative of cosecx from first principle.

Solution : Let y cosecx

and y y cosec(x x)

cosec(x x) cosecx cosec(x x) cosec xy

cosec(x x) cosecx

cosec(x x) cosecxcosec(x x) cosecx

1 1sin(x x) s i n x

cosec(x x) cosecx

sinx sin(x x)cosec(x x) cosecx sin x x sinx

x x2cos x sin2 2

cosec(x x) cosecx sin x x sinx

x 0 x 0

xcos xy 2lim limx cosec(x x) cosecx]

sin x / 2x / 2

[sin(x x).sinx]

or 2dy cosxdx (2 (cosecx)(sinx)

MATHEMATICS

Notes

MODULE - VCalculus

256

Differentiation of Trigonometric Functions121 (cosecx) (cosec x c o t x )

2

Thus,12

d 1cosecx cosecx cosecxcotxdx 2

Example 22.3 Find the derivative of 2sec x from first principle.

Solution : Let 2y sec x

and 2y y sec (x x)

then, 2 2y sec (x x) sec x

2 2

2 2cos x cos (x x)cos (x x)cos x

2 2sin[(x x x]sin[(x x x)]

cos (x x)cos x

2 2sin(2x x)sin xcos (x x)cos x

2 2y sin(2x x)sin xx cos (x x)cos x. x

Now, 2 2x 0 x 0

y sin(2x x)sin xlim limx cos (x x)cos x. x

2 2dy sin2xdx cos xcos x

22 2

2sinxcosx 2tanx.sec xcos xcos x

2secx(sec x.tanx)

= 2sec x (sec x tan x)

CHECK YOUR PROGRESS 22.11. Find the derivative from first principle of the following functions with respect to x :

(a) cosec x (b) cot x (c) cos 2 x

(d) cot 2 x (e) 2cosecx (f) s inx

2. Find the derivative of each of the following functions :

(a) 22sin x (b) 2cosec x (c) 2tan x

MATHEMATICS 257

Notes

MODULE - VCalculus


22.2 DERIVATIVES OF TRIGONOMETRIC FUNCTIONSYou have learnt how we can find the derivative of a trigonometric function from first principleand also how to deal with these functions as a function of a function as shown in the alternativemethod. Now we consider some more examples of these derivatives.


(i) sin 2 x (ii) tan x (iii) 3cosec(5x )

Solution : (i) Let y = sin 2 x,= sin t, where t = 2 x

dy costdt

anddt 2dx

By chain Rule,dy dy dtdx dt dx

, we have

dydx

= cos t (2) = 2. cos t = 2 cos 2 x

Hence,d (sin2x) 2cos2x

dx

(ii) Let y tan x

= tan t where t x

2dy sec tdt

anddt 1dx 2 x

By chain rule,dy dy dtdx dt dx

, we have

dydx

2 1sec t2 x

2sec x2 x

Hence,2d sec xtan x

dx 2 x

Alternatively : Let y tan x

2dy dsec x xdx dx

2sec x2 x

(iii) Let 3y cosec(5x )

3 3 3dy dcosec(5x )cot(5x ) [5x ]dx dx

2 3 315x cosec(5x )cot(5x )

or you may solve it by substituting 3t 5x

MATHEMATICS

Notes

MODULE - VCalculus

258



(i) 4y x sin2x (ii)sinxy

1 cosx

Solution : 4y x sin2x

(i) 4 4dy d dx (sin2x) sin2x (x )dx dx dx

(Using product rule)

4 3x (2cos2x) sin2x(4x )

4 32x cos2x 4x sin2x32x [xcos2x 2sin2x]

(ii)sinxy

1 cosx

2

x x2sin cos2 2

x2cos2

xtan2

2dy x d xsecdx 2 dx 2

21 xsec2 2

Alternatively : You may find the derivative by using quotient rule

Letsin xy

1 cos x

2

d d(1 cosx) (sinx) sin x 1 cos xdy dx dxdx (1 cos x )

2(1 cosx)(cosx) sin x sin x

(1 cos x )

=2 2

2cosx cos x sin x

(1 cos x)

2cosx 1

(1 cos x )

1(1 cos x )

2

1x2cos2

21 xsec2 2

MATHEMATICS 259

Notes

MODULE - VCalculus


Example 22.6 Find the derivative of each of the following functions w.r.t. x :

(i) 2cos x (ii) 3sin x

Solution : (i) Let 2y cos x

2t where t = cos x

dy 2tdt

anddt sin xdx

Using chain rule

dy dy dtdx dt dx

, we have

dydx

= 2 cos x. ( sinx)

2cosxsinx sin2x

(ii) Let 3y sin x

3 1/2 3dy 1 d(sin x) (sin x)dx 2 dx

23

1 3sin x cosx2 sin x

3 sin x cos x2

Thus, 3d 3sin x sin x cos xdx 2

Example 22.7 Finddydx

, when

(i)1 sin xy1 sin x (ii) y a(1 cost),x a(t sint)

Solution : We have,

(i)1 sin xy1 sin x

12dy 1 1 sin x d 1 sinx

dx 2 1 sin x dx 1 sinx

MATHEMATICS

Notes

MODULE - VCalculus

260


21 1 sin x ( cosx)(1 sinx) (1 sinx)(cosx)2 1 sin x (1 sinx)

21 1 sin x 2cosx2 1 sin x (1 sinx)

2

21 sinx 1 sin x1 sinx (1 sinx)

21 sinx 1 sinx 1

1 sinx(1 sinx)

Thus, dy/dx 1

1 sinx

Alternatively, it is more convenient to find the derivative of such square root function byrationalising the denominator.

1 sinx 1 sinxy

1 sinx 1 sinx

21 sinx

1 sin x

1 sinxcos x

secx tanx

22 2

dy sin x 1sec x t a n x sec xdx cos x cos x

2sin x 1 1

1 sin x1 sin x

(ii) x a(t sint), y a(1 cost)

dx dya(1 cost), a(sint)dt dt

Using chain rule,dy dy dtdx dt dx

, we have

dy a(sint)dx a(1 cost)

2

t t2sin cos2 2

t2cos2

ttan2

MATHEMATICS 261

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Example 22.8 Find the derivative of each of the following functions at the indicated points :

(i) 2y sin2x (2x 5) at x2

(ii) 2y cotx sec x 5 at x / 6

Solution :

(i) 2y sin2x (2x 5)

dy d dcos2x (2x) 2(2x 5) (2x 5)dx dx dx

2cos2x 4(2x 5)

At x2

,dy 2cos 4( 5)dx

2 4 204 22

(ii) 2y cotx sec x 5

2dy cosec x 2secx(secxtanx)dx

2 2cosec x 2sec xtanx

At x6

, 2 2dy cosec 2sec tandx 6 6 6

4 14 23 3

843 3

Example 22.9 If sin y = x sin (a+y), prove that2sin a ydy

dx sinaSolution : It is given that

sin y = x sin (a+y) orsinyx

sin(a y) .....(1)

Differentiating w.r.t. x on both sides of (1) we get

2sin(a y)cosy sinycos(a y) dy

1dxsin (a y)

or 2sin(a y y) dy

1dxsin (a y)

MATHEMATICS

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262


or2sin a ydy

dx sina

Example 22.10 If y sinx sinx ....to infinity ,

prove thatdy cosxdx 2y 1

Solution : We are given that

y sinx sinx ...toinfinity

or y sin x y or 2y sinx y

Differentiating with respect to x , we getdy dy2y cosxdx dx

ordy(2y 1) cosxdx

Thus,dy cosxdx 2y 1

CHECK YOUR PROGRESS 22.21. Find the derivative of each of the following functions w.r.t x :

(a) y 3sin4x (b) y cos5x (c) y tan x

(d) y sin x (e) 2y sinx (f) y 2 tan2x

(g) y cot3x (h) y sec10x (i) y cosec2x


(a)secx 1f(x)secx 1

(b)sinx cosxf(x)sinx cosx

(c) f(x) x sinx

(d) 2f(x) 1 x cosx (e) f(x) x cosecx (f) f(x) sin2xcos3x

(g) f(x) sin3x


(a) 3y sin x (b) 2y cos x (c) 4y tan x

(d) 4y cot x (e) 5y sec x (f) 3y cosec x

(g) y sec x (h)secx tanx

ysecx tanx

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Differentiation of Trigonometric Functions4. Find the derivative of the following functions at the indicated points :

(a) y cos(2x / 2),x3

(b)1 sinxy , x

cosx 4

5. If y tanx tanx tanx ..., to infinity

Show that 2dy(2y 1) sec xdx

.

6. If cosy xcos(a y),

prove that 2dy cos (a y)

dx sina.

22.3 DERIVATIVES OF INVERSE TRIGONOMETRICFUNCTIONS FROM FIRST PRINCIPLE

We now find derivatives of standard inverse trignometric functions 1sin x, 1cos x, 1tan x, byfirst principle.

(i) We will show that by first principle the derivative of 1sin x w.r.t. x is given by

12

d 1(sin x)dx 1 x

Let 1y sin x . Then x = sin y and so x x sin(y y)As x 0, y 0 .Now, x sin(y y) sin y

sin(y y) s iny1x

[On dividing both sides by x ]

or sin(y y) sin y y1y x

y 0 x 0

sin(y y) siny y1 lim limy x [ y 0 when x 0]∵

y 0

1 12cos y y sin ydy2 2lim

y dx

dycosydx

2 2

dy 1 1 1dx cos y 1 sin y 1 x

MATHEMATICS

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264


12

d 1sin x

dx 1 x

(ii)1

2

d 1cos xdx 1 x

For proof proceed exactly as in the case of 1sin x .

(iii) Now we show that,

12

d 1tan xdx 1 x

Let 1y tan x .Then x tany and so x x tan(y y)

As x 0, also y 0Now, x tan(y y) tany

tan(y y) tany y1y x

y 0 x 0

tan(y y) tany y1 lim limy x y 0 when x 0∵

y 0

dysin(y y) sinylim ycos(y y) cosy dx

y 0

dy sin(y y)cosy cos(y y)sinylimdx y.cos(y y)cosy

y 0

dy sin(y y y)limdx y.cos(y y)cosy

y 0

dy sin y 1lim

dx y cos(y y)cosy

22

dy 1 dy sec ydx dxcos y

2 2 2dy 1 1 1dx sec y 1 tan y 1 x

12

d 1tan xdx 1 x

(iv)1

2d 1cot x

dx 1 x

For proof proceed exactly as in the case of 1tan x .

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Notes

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(v) We have by first principle 12

d 1(sec x)dx x x 1

Let 1y sec x . Then x = sec y and so x x sec(y y).

As x 0,also y 0 .Now x sec(y y) secy.

sec(y y) secy y1y x

y 0 x 0

sec(y y) secy y1 lim limy x y 0 when x 0∵

y 0

1 12sin y y sin ydy 2 2limdx y.cosycos y y

y 0

1 1sin y y sin ydy 2 2lim 1dx cosycos y y y

2

=dy siny dy secytanydx cosycosy dx

2 2

dy 1 1 1dx secytany secy sec y 1 x x 1

12

d 1sec xdx x x 1

(vi) 12

d 1cosec x .dx x x 1

For proof proceed as in the case of 1sec x.

Example 22.11 Find derivative of 1 2sin x from first principle.

Solution : Let 1 2y sin x

2x siny

Now, 2x x sin(y y)

2 2x x x sin y y sinyx x

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266


2 2

x 0 y 0 x 0

y y2cos y sinx x x y2 2lim lim limy(x x) x 2 x2

dy2x cosydx

2 4dy 2x 2x 2xdx cosy 1 sin y 1 x

.

Example 22.12 Find derivative of 1sin x w.r.t. x by delta method.

Solution : Let 1y sin x

siny x ..(1)

Also sin(y y) x x ..(2)From (1) and (2), we get

sin(y y) siny x x x

orx x x x x xy y

2cos y sin2 2 x x x

xx x x

y y2cos y sin12 2

x x x x

or

ysiny y 12cos y yx 2 x x x

2

x 0 y 0 y 0

ysiny y 2lim lim cos y lim yx 2

2

x 0

1limx x x y 0 as x 0∵

ordy 1c o s ydx 2 x or

2dy 1 1 1dx 2 x cosy 2 x 1 x2 x 1 sin y

dy 1dx 2 x 1 x

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Notes

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CHECK YOUR PROGRESS 22.31. Find by first principle that derivative of each of the following :

(i) 1 2cos x (ii)1cos x

x(iii) 1cos x

(iv) 1 2tan x (v)1tan x

x(vi) 1tan x

22.4 DERIVATIVES OF INVERSE TRIGONOMETRIC FUNCTIONSIn the previous section, we have learnt to find derivatives of inverse trignometric functions byfirst principle. Now we learn to find derivatives of inverse trigonometric functions by alternative

methods. We start with standard inverse trignometric functions 1 1sin x,cos x,....

(i) Derivative of 1sin x

Solution : Let 1y sin x x = sin y (i)

Differentiating w.r.t. ydx cosydy

2dx 1 sin ydy ...[Using (i) ]

or 21 1dx 1 sin ydy

2dy 1dx 1 sin y 2

1

1 x

dy 1dxdxdy

∵

Hence,1

2d 1[sin x]

dx 1 xSimilarly we can show that

12

d 1[cos x]dx 1 x

(ii) Derivative of 1tan x

Solution : Let 1tan x y

x tany

Differentiating w.r.t. y,2dx sec y

dy

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268


and 2dy 1dx sec y

21

1 tan y [∵ We have written tan y in terms of x]

21

1 x

Hence, 12

d 1tan xdx 1 x

Similarly, 12

d 1cot xdx 1 x

.

(iii) Derivative of 1sec x

Solution : Let 1sec x y

x = sec y and dx secytanydy

dy 1dx sec y tan y

21

sec y [ sec y 1]

21

sec y sec y 1

21

| x | sec y 1

Note : (i) When x >1, sec y is + ve and tan y is + ve, y 0,2

(ii) When x 1, sec y is ve and tan y is ve, y ,2

Hence,1

2

d 1(sec x)dx | x | x 1

Similarly1

2

d 1(cosec x)dx | x | x 1


(i) 1sin x (ii) 1 2cos x (iii) 1 2(cosec x)

Solution :

(i) Let 1y sin x

MATHEMATICS 269

Notes

MODULE - VCalculus


2

dy 1 d xdx dx

1 x

1 / 21 1 x21 x

12 x 1 x

1d 1sin xdx 2 x 1 x

(ii) Let 1 2y cos x

22 2

dy 1 d (x )dx dx1 (x )

41 (2x)

1 x

1 24

d 2xcos xdx 1 x

(iii) Let 1 2y (cosec x)

1 1dy d2(cosec x) cosec xdx dx

12

12(cosec x)| x | x 1

1

2

2cosec x

| x | x 1

11 2

2

d 2cosec x(cosec x)dx | x | x 1


(i) 1 cosxtan1 sinx

(ii) 1sin(2sin x)

Solution :

Let (i) 1 cosxy tan1 sinx

1sin x

2tan1 cos x

2

MATHEMATICS

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270


1 xtan tan

4 2

x4 2

dy 1/2dx

(ii) 1y sin(2sin x)

Let 1y sin(2sin x)

1 1dy dcos(2sin x) (2sin x)dx dx

12

dy 2cos(2sin x)dx 1 x

=1

22cos(2sin x)

1 x

Example 22.15 Show that the derivative of 1 12 2

2x 2xtan w.r.t sin1 x 1 x

is 1.

Solution : Let 1 12 2

2x 2xy tan and z sin1 x 1 x

Let x tan

12

2tany tan1 tan

and 12

2tanz sin1 tan

1tan (tan2 ) and 1z sin (sin2 ) = 2 and z = 2

dy 2d

anddz 2d

dy dy ddx d dz

12 12

(By chain rule)


Find the derivative of each of the following functions w.r.t. x and express the resultin the simplest form (1-3) :

1. (a) 1 2sin x (b) 1 xcos2

(c) 1 1cosx

2. (a) 1tan (cosecx cotx) (b) 1cot (secx tanx) (c)1 cosx sinxtan

cosx sinx

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Notes

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3. (a) 1sin(cos x) (b) 1sec(tan x) (c) 1 2sin (1 2x )

(d) 1 3cos (4x 3x) (e) 1 2cot 1 x x

4. Find the derivative of :1

1tan x

1 tan x w.r.t. 1tan x .

22.5 SECOND ORDER DERIVATIVESWe know that the second order derivative of a function is the derivative of the first derivative ofthat function. In this section, we shall find the second order derivatives of trigonometric andinverse trigonometric functions. In the process, we shall be using product rule, quotient rule andchain rule.

Let us take some examples.

Example 22.16 Find the second order derivative of(i) s inx (ii) xcosx (iii) 1cos x

Solution : (i) Let y = sin x

Differentiating w.r.t. x both sides, we getdy cosxdx

Differentiating w.r.t x both sides again, we get2

2d y d (cosx) sinx

dxdx

2

2d y sinxdx

(ii) Let y = x cos xDifferentiating w.r.t. x both sides, we get

dy x( sinx) cosx.1dx

dy xsinx cosxdx

Differentiating w.r.t. x both sides again, we get2

2d y d xsinx cosx

dxdx

x.cosx sinx sinx

x.cosx 2sinx2

2d y x.cosx 2sinxdx

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272


(iii) Let 1y cos x

Differentiating w.r.t. x both sides, we get1

2 21 / 22 2

dy 1 1 1 xdx 1 x 1 x

Differentiating w.r.t. x both sides, we get2 3 / 22

2d y 1 1 x 2x

2dx

3 / 22

x

1 x

2

2 3 / 22

d y xdx 1 x

Example 22.17 If 1y sin x , show that 22 11 x y xy 0, where 2y and 1y respectively

denote the second and first, order derivatives of y w.r.t. x.

Solution : We have, 1y sin x

Differentiating w.r.t. x both sides, we get

2dy 1dx 1 x

or2

2dy 1dx 1 x

(squaring both sides)

or 2 211 x y 0

Differentiating w.r.t. x both sides, we get

2 21 1 1

d1 x 2y y 2x y 0dx

or 2 21 2 11 x 2y y 2 x y 0

or 22 11 x y x y 0

CHECK YOUR PROGRESS 22.51. Find the second order derivative of each of the following :

(a) sin(cosx) (b) 2 1x tan x

MATHEMATICS 273

Notes

MODULE - VCalculus


2. If211y sin x

2, show that 2

2 11 x y xy 1 .

3. If y sin(sinx) , prove that 2

22

d y dytanx ycos x 0dxdx

.

4. If y x tanx, show that 2

22

d ycos x 2y 2x 0dx

● (i)d (sinx) cosx

dx(ii)

d (cosx) sinxdx

(iii) 2d (tanx) sec xdx

(iv) 2d (cotx) cosec xdx

(v)d (secx) secxtanx

dx(vi)

d (cosecx) cosecxcotxdx

● If u is a derivabale function of x, then

(i)d du(sinu) cosu

dx dx(ii)

d du(cosu) sinudx dx

(iii) 2d du(tanu) sec udx dx

(iv) 2d du(cotu) cosec udx dx

(v)d du(secu) secutanu

dx dx(vi)

d du(coseu) cosec u c o t udx dx

● (i)1

2d 1(sin x)

dx 1 x(ii)

12

d 1(cos x)dx 1 x

(iii) 12

d 1(tan x)dx 1 x

(iv) 12

d 1(cot x)dx 1 x

(v)1

2

d 1(sec x)dx | x | x 1

(vi)1

2

d 1(cosec x)dx | x | x 1

● If u is a derivable function of x, then

(i)1

2d 1 du(sin u)

dx dx1 u(ii)

12

d 1 du(cos u)dx dx1 u

(iii) 12

d 1 du(tan u)dx dx1 u

(iv) 12

d 1 du(cot u)dx dx1 u

(v)1

2

d 1 du(sec u)dx dx| u | u 1

(vi)1

2

d 1 du(cosec u)dx dx| u | u 1

The second order derivative of a trignometric function is the derivative of their first orderderivatives.

LET US SUM UP

MATHEMATICS

Notes

MODULE - VCalculus

274



TERMINAL EXERCISE

1. If 3 2 xy x tan2

, find dydx

.

2. Evaluate, 4 4d sin x cos xdx

at x2

and 0.

3. If 223

5xy cos (2x 1)(1 x)

, find dydx

.

4. If 1 1x 1 x 1y sec sin

x 1 x 1, then show that

dydx

= 0

5. If 3 3x acos , y asin , then find2dy1

dx.

6. If y x x x .... , find dydx

.

7. Find the derivative of 1sin x w.r.t. 1 2cos 1 x

8. If y cos(cosx) , prove that

22

2d y dycotx y.sin x 0

dxdx.

9. If 1y tan x show that

22 1(1 x) y 2xy 0 .

10. If 1 2y (cos x ) , show that2

2 1(1 x )y xy 2 0 .


MATHEMATICS 275

Notes

MODULE - VCalculus


ANSWERS

CHECK YOUR PROGRESS 22.1(1) (a) cosecxco tx (b) 2cosec x (c) 2sin2x

(d) 22cosec 2x (e) 2 22xcosecx cotx (f)cosx

2 sinx

2. (a)2sin2x (b) 22cosec xcotx (c) 22tanxsec x


1. (a)12cos4x (b) 5sin5x (c)2sec x

2 x(d)

cos x2 x

(e) 22 x c o s x (f) 22 2 sec 2x (g) 23 cosec 3x(h)10sec10xtan10x (i) 2cosec2xcot2x

2. (a) 22secxtanx(secx 1) (b) 2

2(sinx cosx) (c) xcosx sinx

(d) 22xcosx (1 x )sinx

(e)cosecx(1 xcotx) (f) 2cos2xcos3x 3 s i n 2 x s i n 3 x (g)3cos3x

2 sin3x

3. (a) 23sin xcosx (b) sin2x (c) 3 24tan xsec x (d) 3 24cot xcosec x

(e) 55sec x t a n x (f) 33cosec x cotx (g) sec x tan x2 x

(h) secx secx tanx4. (a) 1 (b) 2 2


1. (i)4

2x

1 x(ii)

1

221 cos x

xx 1 x(iii) 1

2

1

2x (1 x)

(iv) 42x

1 x(v)

1

2 21 tan x

x(1 x ) x(vi)

12

1

2x 1 x


1. (a) 42x

1 x(b) 2

1

4 x(c) 2

1

x x 1

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276


2. (a)12

(b)12

(c) 1

3. (a)1

2

cos cos x

1 x(b) 1

2x sec tan x

1 x

(c) 22

1 x(d) 2

3

1 x(e) 2

12(1 x )

4. 21

1

1 tan x


1. (a) 2cosxcos(cosx) sin xsin(cosx)

(b)2

12 2

2x(2 x ) 2tan x(1 x )

TERMINAL EXERCISE

1. 3 2 2 2x x xx tan sec 3x tan2 2 2

2. 0, 0

3. 53

5(3 x) 2sin(4x 2)

3(1 x)

5. |sec |

6.1

2y 1 7. 21

2 1 x

MATHEMATICS 277

Notes

MODULE - VCalculus

Differentiation of Exponential and Logarithmic Functions

23

DIFFERENTIATION OF EXPONENTIALAND LOGARITHMIC FUNCTIONS

We are aware that population generally grows but in some cases decay also. There are manyother areas where growth and decay are continuous in nature. Examples from the fields ofEconomics, Agriculture and Business can be cited, where growth and decay are continuous.Let us consider an example of bacteria growth. If there are 10,00,000 bacteria at present andsay they are doubled in number after 10 hours, we are interested in knowing as to after howmuch time these bacteria will be 30,00,000 in number and so on.Answers to the growth problem does not come from addition (repeated or otherwise), ormultiplication by a fixed number. In fact Mathematics has a tool known as exponential functionthat helps us to find growth and decay in such cases. Exponential function is inverse of logarithmicfunction.In this lesson, we propose to work with this tool and find the rules governing their derivatives.

OBJECTIVESAfter studying this lesson, you will be able to :● define and find the derivatives of exponential and logarithmic functions;● find the derivatives of functions expressed as a combination of algebraic, trigonometric,

exponential and logarithmic functions; and● find second order derivative of a function.

EXPECTED BACKGROUND KNOWLEDGE� Application of the following standard limits :

(i)n

n

1lim 1 e

n(ii)

1n

nlim 1 n e

(iii)x

x

e 1lim 1

x(iv)

x

ex

a 1lim log a

x

(v)h

h 0

e 1lim 1h

� Definition of derivative and rules for finding derivatives of functions.

MATHEMATICS

Notes

MODULE - VCalculus

278


23.1 DERIVATIVE OF EXPONENTIAL FUNCTIONSLet xy e be an exponential function. .....(i)

x xy y e (Corresponding small increments) .....(ii)

From (i) and (ii), we havex x xy e e

Dividing both sides by x and taking the limit as x 0

xx

x 0 x 0

y e 1lim lim ex x

x xdy e .1 edx

Thus, we have x xd e edx

.

Working rule : x x xd de e x edx dx

Next, let ax by e .

Then a(x x) by y e[ x and y are corresponding small increments]

a x x b ax by e e

ax b a xe e 1

a xax b

e 1y ex x

a xax b e 1a e

a x[Multiply and divide by a]

Taking limit as x 0 , we havea x

ax bx 0 x 0

y e 1lim a e limx a x

or ax bdy a e 1dx

x

x 0

e 1lim 1x

ax bae

MATHEMATICS 279

Notes

MODULE - VCalculus


Working rule :

ax b ax b ax bd de e ax b e adx dx

ax b ax bd e aedx

Example 23.1 Find the derivative of each of the follwoing functions :

(i) 5xe (ii) axe (iii)3x2e

Soution : (i) Let 5xy e .

Then ty e where 5 x = t

tdy edt

anddt5dx

We know that, dy dy dtdx dt dx

te 5 5x5e

Alternatively 5x 5x 5xd de e 5x e 5dx dx

5x5e

(ii) Let axy e

Then ty e when t = ax

tdy edt

and dt adx

We know that, tdy dy dt e adx dt dx

Thus, axdy a edx

(iii) Let3x2y e

3 x2dy d 3e x

dt dx 2

Thus,3x2dy 3 e

dt 2


(i) xy e 2cosx (ii)2x x5y e 2sinx e 2e

3

MATHEMATICS

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MODULE - VCalculus

280


Solution : (i) xy e 2cosx

xdy d de 2 cosxdx dx dx

xe 2sinx

(ii)2x x5y e 2sinx e 2e

32x 2 xdy d 5e x 2cosx e 0

dx dx 3 [Since e is constant]

2x x52xe 2cosx e3

Example 23.3 Find dydx

, when

(i) xcosxy e (ii) x1y ex

(iii)1 x1 xy e

Solution : (i) xcosxy e

x c o s xdy de x c o s xdx dx

x c o s xdy d de x cosx cosx (x)

dx dx dxxcosxe xsinx cosx

(ii) x1y ex

x xdy d 1 1 de e

dx dx x x dx [Using product rule]

x x21 1e e

xxx x

2 2e e[ 1 x] [x 1]x x

(iii)1 x1 xy e1 x1 xdy d 1 xe

dx dx 1 x

1 x1 x

21.(1 x) (1 x).1e

(1 x)

1 x1 x

22e

(1 x)

1 x1 x

22 e

(1 x)

MATHEMATICS 281

Notes

MODULE - VCalculus



(i) sinx xe sine (ii) axe cos bx c

Solution : sinx xy e sine

sinx x x sinxdy d de sine sine edx dx dx

sinx x x x sinxd de cose e sine e sinxdx dx

sinx x x x sinxe cose e sine e cosxsinx x x xe [e cose sine cosx]

(ii) axy e cos(bx c)

ax axdy d de cos(bx c) cos bx c edx dx dx

ax axd de [ sin(bx c)] (bx c) cos(bx c)e (ax)dx dx

ax axe sin(bx c) b cos(bx c)e .a

axe [ bsin(bx c) acos(bx c)]


, ifaxey

sin(bx c)

Solution :ax ax

2

d dsin(bx c) e e [sin(bx c)]dy dx dxdx sin (bx c)

ax ax

2sin(bx c).e .a e cos(bx c).b

sin (bx c)

ax

2e [asin(bx c) bcos(bx c)]

sin (bx c)

CHECK YOUR PROGRESS 23.11. Find the derivative of each of the following functions :

(a) 5xe (b) 7x 4e (c) 2xe (d)7 x

2e (e)2x 2xe

2. Finddydx

, if

MATHEMATICS

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282


(a) x1y e 5e3

(b) x1y tanx 2sinx 3cosx e2

(c) xy 5sinx 2e (d) x xy e e


(a) x 1f(x) e (b) cotxf(x) e

(c) 2xsin xf(x) e (d) 2xsec xf(x) e

4. Find the derivative of each of the following functions :(a) xf(x) (x 1)e (b) 2x 2f(x) e sin x

5. Finddydx

, if

(a)2x

2

ey

x 1(b)

2xe c o s xyx sin x

23.2 DERIVATIVE OF LOGARITHMIC FUNCTIONSWe first consider logarithmic functionLet y l o g x .....(i)

y y log x x .....(ii)

( x and y are corresponding small increments in x and y)From (i) and (ii), we get

y log x x l o g x

x xlogx

y 1 xlog 1

x x x

1 x xlog 1

x x x [Multiply and divide by x]

xx1 xlog 1

x x

Taking limits of both sides, as x 0 , we get

xx

x 0 x 0

y 1 xlim lim log 1x x x

MATHEMATICS 283

Notes

MODULE - VCalculus


xx

x 0

dy 1 xlog lim 1

dx x x

1 logex

xx

x 0

xlim 1 ex

∵

1x

Thus,d 1(logx)

dx xNext, we consider logarithmic function

y log(ax b) ...(i)

y y log[a(x x) b] ...(ii)[ x and y are corresponding small increments]

From (i) and (ii), we gety log[a(x x) b] log(ax b)

a(x x) blogax b

(ax b) a xlogax b

a xlog 1

ax b

y 1 a xlog 1

x x ax b

a ax b a xlog 1

ax b a x ax ba

Multiply and divide by ax+b

ax ba xa a x

log 1ax b ax b

Taking limits on both sides as x 0ax ba x

x 0 x 0

y a a xlim lim log 1x ax b ax b

ordy a logedx ax b

1x

x 0lim 1 x e∵

or,dy adx ax b

MATHEMATICS

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284

Differentiation of Exponential and Logarithmic FunctionsWorking rule :

d 1 dlog(ax b) (ax b)dx ax b dx

1 aaax b ax b

Example 23.6 Find the derivative of each of the functions given below :

(i) 5y logx (ii) y log x (iii) 3y logx

Solution : (i) 5y logx = 5 log x

dy 1 55dx x x

(ii) y log x12logx or

1y logx2

dy 1 1 1dx 2 x 2x

(iii) 3y logx

3y t , when t logx

2dy 3tdt

anddt 1dx x

We know that, 2dy dy dt 13tdx dt dx x

2dy 13(logx)dx x

2dy 3 (logx)dx x

Example 23.7 Find,dydx

if

(i) 3y x logx (ii) xy e logx

Solution :(i) 3y x logx

3 3dy d dlogx (x ) x (logx)dx dx dx

[Using Product rule]

2 3 13x logx xx

2x 3logx 1

MATHEMATICS 285

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(ii) xy e logx

x xdy d de (logx) logx edx dx dx

x x1e e logxx

x 1e logx

x


(i) log tan x (ii) log [cos (log x )]Solution : (i) Let y = log tan x

dy 1 d (tanx)dx tanx dx

21 sec xtanx

2cosx 1s i n x cos x

= cosec x.sec x(ii) Let y = log [cos (log x)]

dy 1 d [cos(logx)]dx cos(logx) dx

1 dsinlogx logxcos logx dx

sin(logx) 1cos(logx) x

1 tan(logx)x


, if y = log(sec x + tan x)

Solution : y = log (sec x + tax x )

dy 1 d secx t a n xdx secx t a n x dx

21 sec x tanx sec xsecx t a n x

1 secx sec x t a n xsecx t a n x

MATHEMATICS

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286


secx(tanx secx)secx t a n x

= sec x

Example 23.10 Finddydx

, if

12 2 2

33 4

(4x 1)(1 x )y

x (x 7)

Solution : Although, you can find the derivative directly using quotient rule (and product rule)but if you take logarithm on both sides, the product changes to addition and division changes tosubtraction. This simplifies the process:

12 2 2

33 4

(4x 1)(1 x )y

x (x 7)

Taking logarithm on both sides, we get

12 2 2

33 4

4x 1 1 xlogy log

x x 7

or 2 21 3logy log 4x 1 log 1 x 3logx log(x 7)2 4

[ Using log properties]Now, taking derivative on both sides, we get

2 2d 1 1 3 3 1(logy) 8x 2x

dx x 4 x 74x 1 2(1 x )

2 21 dy 8x x 3 3y dx x 4 x 74x 1 1 x

2 2dy 8x x 3 3

ydx x 4(x 7)4x 1 1 x

2 2

3 2 23 4

(4x 1) 1 x 8x x 3 3x 4(x 7)4x 1 1 x

x (x 7)

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Notes

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CHECK YOUR PROGRESS 23.21. Find the derivative of each the functions given below:

(a)f(x)=5sinx� 2 log x (b) f (x) = log cos x

2. Finddydx

, if

(a) 2xy e logx (b)2xey

logx


(a) y = log ( sin log x ) (b) y = log tan x

4 2

(c)a btanxy loga btanx

(d) y = log (log x )

4. Finddydx

, if

(a)2 11 3

23 72 2y (1 x) (2 x) (x 5) (x 9) (b)

32

5 124 4

x(1 2x)y

(3 4x) (3 7x )

23.3 DERIVATIVE OF LOGARITHMIC FUNCTION (CONTINUED)

We know that derivative of the function nx w.r.t. x is n n 1x , where n is a constant. This rule isnot applicable, when exponent is a variable. In such cases we take logarithm of the function andthen find its derivative.

Therefore, this process is useful, when the given function is of the type g xf x . For example,

x xa ,x etc.

Note : Here f(x) may be constant.

Derivative of ax w.r.t. x

Let xy a , a > 0

Taking log on both sides, we getxlogy loga xloga [ nlogm nlogm]

d dlogy (xloga)dx dx

or1 dy dloga xy dx dx

ordy ylogadx

MATHEMATICS

Notes

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288

Differentiation of Exponential and Logarithmic Functionsxa loga

Thus, x xd a a logadx

, a > 0


(i) xy x (ii) sinxy x

Solution : (i) xy x

Taking logrithms on both sides, we getlogy xlogx

Taking derivative on both sides, we get

1 dy d dlogx (x) x (logx)y dx dx dx [Using product rule]

1 dy 11 logx xy dx x

= log x + 1

dy y[logx 1]dx

Thus, xdy x (logx 1)dx

(ii) sinxy x

Taking logarithm on both sides, we get log y = sin x log x

1 dy d (sinxlogx)y dx dx

or1 dy 1cosx.logx sinxy dx x

ordy sinx

y cosxlogxdx x

Thus, sinxdy sinxx cosxlogx

dx x

Example 23.12 Find the derivative, ifs i n xx 1y logx sin x

Solution : Here taking logarithm on both sides will not help us as we cannot put

MATHEMATICS 289

Notes

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Differentiation of Exponential and Logarithmic Functionsx 1 sinx(logx) (sin x) in simpler form. So we put

xu (logx) and 1 sinxv (sin x)Then, y u v

dy du dvdx dx dx

......(i)

Now xu (logx)

Taking log on both sides, we havexlogu log(logx)

logu xlog(logx) nlog m n l o g m∵

Now, finding the derivative on both sides, we get1 du 1 11 log(logx) xu dx logx x

Thus,du 1

u log(logx)dx logx

Thus, xdu 1(logx) log(logx)

dx logx .....(ii)

Also, 1 sinxv (sin x)

1logv sinxlog(sin x)

Taking derivative on both sides, we have

1d d(logv) [sinxlog(sin x)]dx dx

11 2

1 dv 1 1sinx cosx log sin xv dx sin x 1 x

or,1

1 2

dv sinxv cosx logsin xdx sin x 1 x

sinx1 11 2

sinxsin x cosxlog sin xsin x 1 x

....(iii)

From (i), (ii) and (iii), we havesinxx 1 1

1 2

dy 1 sinxlogx log logx sin x cosxlogsin xdx logx sin x 1 x

Example 23.13 If y x yx e , prove that

2dy logxdx 1 l o g x

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290


Solution : It is given that y x yx e ...(i)

Taking logarithm on both sides, we getylogx (x y)loge

(x y)

or y(1 logx) x loge 1∵

orxy

1 logx (ii)

Taking derivative with respect to x on both sides of (ii), we get

2

11 logx 1 xdy xdx 1 l o g x

2 21 logx 1 l o g x

1 logx 1 logx

Example 23.14 Find , dydx

if

x 1 1e logy sin x sin y

Solution : We are given thatx 1 1e logy sin x sin y

Taking derivative with respect to x of both sides, we get

x x2 2

1 dy 1 1 dye e l o g yy d x dx1 x 1 y

orx

x2 2

e 1 dy 1 e l o g yy dx1 y 1 x

or2 x 2

x 2 2

y 1 y 1 e 1 x l o g ydydx e 1 y y 1 x


, if cosx ......co sxy cosx

Solution : We are given thatcosx ......cosx yy cosx c o s x

Taking logarithm on both sides, we get

MATHEMATICS 291

Notes

MODULE - VCalculus


l ogy y logcosx

Differentiating (i) w.r.t.x, we get1 dy 1 dyy s i n x log c o s xy dx cos x dx

or1 dylog cosx y t a n xy dx

or 2dy1 ylog cosx y t a n xdx

or2dy y t a n x

dx 1 ylog c o s x

CHECK YOUR PROGRESS 23.31. Find the derivative with respect to x of each the following functions :

(a) xy 5 (b) x xy 3 4 (c) xy sin(5 )

2. Finddydx

, if

(a) 2xy x (b) logxy (cosx) (c) sinxy (logx)

(d) xy (tanx) (e) 22 xy (1 x ) (f) 2(x sinx)y x


(a) cotx xy (tanx) (cotx) (b) 1logx sin xy x (sinx)

(c) tanx cosxy x (sinx) (d) 2x logxy (x) (logx)

4. Ifsinx. . . . . .s i n xy s i n x , show that

2dy y cot xdx 1 ylog s i n x

5. If y logx logx logx ...... , show that

dy 1dx x 2x 1

MATHEMATICS

Notes

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292


23.4 SECOND ORDER DERIVATIVESIn the previous lesson we found the derivatives of second order of trigonometric and inversetrigonometric functions by using the formulae for the derivatives of trigonometric and inversetrigonometric functions, various laws of derivatives, including chain rule, and power rule discussedearlier in lesson 21. In a similar manner, we will discuss second order derivative of exponentialand logarithmic functions :

Example 23.16 Find the second order derivative of each of the following :

(i) xe (ii) cos (logx) (iii) xx

Solution : (i) Let xy e

Taking derivative w.r.t. x on both sides, we get xdy edx

Taking derivative w.r.t. x on both sides, we get 2

x x2

d y d (e ) edxdx

2x

2d y edx

(ii) Let y = cos (log x)Taking derivative w.r.t x on both sides, we get

sin l o g xdy 1sin l o g xdx x x

Taking derivative w.r.t. x on both sides, we get2

2sin l o g xd y d

dx xdx

or2

1x cos logx sin l o g xx

x2

2 2sin logx cos l o g xd y

dx x(iii) Let xy xTaking logarithm on both sides, we get

log y = x log x ....(i)Taking derivative w.r.t. x of both sides, we get

1 dy 1x logx 1 logxy dx x

ordy y(1 logx)dx

....(ii)

Taking derivative w.r.t. x on both sides we get

MATHEMATICS 293

Notes

MODULE - VCalculus


2

2d y d y(1 logx)

dxdx

=1 dyy (1 logx)x dx

...(iii)

y (1 logx)y(1 logx)x

2y (1 logx) yx

(Using (ii))

21y (1 logx)

x

22x

2d y 1x 1 l o g x

xdx

Example 23.17 If 1acos xy e , show that

22 2

2d y dy1 x x a y 0

dxdx.

Solution : We have, 1acos xy e .....(i)

1acos x2

dy aedx 1 x

2

ay

1 xUsing (i)

or2 2 2

2dy a ydx 1 x

22 2 2dy 1 x a y 0

dx.....(ii)

Taking derivative of both sides of (ii), we get

2 22 2

2dy dy d y dy2x 2 1 x a 2y 0dx dx dxdx

or2

2 22

d y dy1 x x a y 0dxdx

[Dividing through out bydy2dx

]

MATHEMATICS

Notes

MODULE - VCalculus

294


CHECK YOUR PROGRESS 23.41. Find the second order derivative of each of the following :

(a) 4 5xx e (b) 5xtan e (c)l o g x

x

2. If y acos logx bsin l o g x , show that

22

2d y dyx x y 0

dxdx

3. If1tan xy e , prove that

22

2d y dy1 x 2x 1 0

dxdx

� (i) x xd (e ) edx

(ii) x xd (a ) a loga ; a 0dx

� If is a derivable function of x, then

(i)d d(e ) e

dx dx(ii)

d d(a ) a loga ;a 0dx dx

(iii) ax b ax b ax bd (e ) e a aedx

� (i)d 1(logx)

dx x(ii)

d 1 d(log )dx dx , if is a derivable function of x.

� (iii)d 1 alog(ax b) a

dx ax b ax b


LET US SUM UP


MATHEMATICS 295

Notes

MODULE - VCalculus


TERMINAL EXERCISE


(a) x x(x ) (b)xx

x

2. Finddydx

, if

(a) x logs inxy a (b)1cos xy (sinx)

(c)2x1y 1

x(d)

34x x 4

y log ex 4


(a) 2x xf(x) cosxlog(x)e x (b) 1 2 sinx 2xf(x) (sin x) x e


(a) logx s inxy (tanx) (cosx) (b) tanx cosxy x (sinx)

5. Finddydx

, if

(a) If4

2x x 6y(3x 5)

(b) Ifx x

x xe ey

(e e )

6. Finddydx

, if

(a) If x ay a x (b) 2x 2xy 7


(a) 2 2xy x e cos3x (b)x2 c o t xy

x

8. Ifx.......xxy x , prove that

2dy yxdx 1 y l o g x

MATHEMATICS

Notes

MODULE - VCalculus

296


ANSWERS


1. (a) 5x5e (b) 7x 47e (c) 2x2e (d)7 x27 e

2(e)

2x 2x2 x 1 e

2. (a) x1e3

(b) 2 x1sec x 2cosx 3s inx e2

(c) x5cosx 2e (d) x xe e

3. (a)x 1e

2 x 1(b)

2c o t x cosec xe

2 c o t x

(c)2xsin xe [ s inx 2xcosx]s inx

(d)2xsec x 2 2e sec x 2xsec x t a n x

4. (a) xxe (b) 2x2e sinx(sinx cosx)

5. (a)2

2x2 3 / 2

2x x 2 e(x 1)

(b)2x 2

2e [(2x 1)cotx xcosec x]

x


1. (a) 25cosxx

(b) tanx

2. (a)2x 1

e 2xlogxx

(b)2 2x

22x logx 1.e

x(logx)

3. (a)cot(logx)

x(b) secx

(c)2

2 2 22absec x

a b tan x(d)

1xlogx

4. (a)2 11 3

23 72 2(1 x) (2 x) (x 5) (x 9) 21 2 2x 3

2(1 x) 3(2 x) 2(x 9)7(x 5)

(b)

32

5 124 4

x(1 2x)

(3 4x) (3 7x )2

1 3 5 7x2x 1 2x 3 4x 2(3 7x )

MATHEMATICS 297

Notes

MODULE - VCalculus


CHECK YOUR PROGRESS 23.31. (a) x5 log5 (b) x x3 log3 4 log4 (c) x xcos5 5 log5

2. (a) 2x2x (1 logx) (b) logx logcosx(cosx) tanxlogx

x

(c) s i n x s i n xlogx cosxlog logxx l o g x

(d) x xtanx l o g t a n x

s i n x c o s x

(e)32x 2

2x

(1 x) 2xlog(1 x ) 21 x

(f)22(x sinx) x sinx

x (2x cosx)logxx

3. (a) cotx x2 2cosec x 1 logtanx tanx logcotx xcosec xtanx cotx

(b)1sin xlogx 1 1

2

logsinx2x logx sinx cotxsin x1 x

(c) cosxtanx 2tanxx sec xlogx sinx cosxcotx sinxlogsinxx

(d)2 logxx 1 log logx

x x 1 2logx logxx


1. (a) 5x 4 3 2e 25x 40x 12x (b) 5x 2 x 5x 5x25e sec e 1 2e tane

(c) 32 logx 3

xTERMINAL EXERCISEs to Terminal uestions

1. (a) x x(x ) [x 2xlogx] (b) x(x) x 1 xx [x logx(logx 1)x ]

2. (a) xlogsinxa [logsinx xcotx]loga

(b)1cos x 1

2logs inxs inx cos x c o t x

1 x

MATHEMATICS

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298


(c)2x1

1x

1 12xlog x1x 1x

(d) 3 314(x 4) 4(x 4)

3. (a) 2x x 1cosxlog(x)e x tanx 2x 1 logx

xlogx

(b) 1 2 sinx 2x2 1

2 sinx(sin x) .x e cosxlogx 2

x1 x sin x

4. (a) logx 1(tanx) 2cosec2x logx logtanx

x

sinx(cosx) [ sinxtanx cosxlog(cosx)]

(b) tanx 2tanxx sec xlogx

x+ cosx(sinx) [cotxcosx sinxlogsinx)]

5. (a)4

2x x 6(3x 5)

4 1 6x 2(x 6) (3x 5) (b)

2x

2x 24e

(e 1)

6. (a) x a 1ea .x [a xlog a] (b) 2x 2x

e7 (2x 2)log 7

7. (a) 2 2x 2x e cos3x 2 3tan3x

x

(b)x2 cotx 1log2 2cosec2x

2xx

MATHEMATICS 299

Notes

MODULE - VCalculus

Tangents and Normals

24

TANGENTS AND NORMALS

In the previous lesson, you have learnt that the slope of a line is the tangent of the angle whichthis line makes with the positive direction of x-axis. It is denoted by the letter 'm'. Thus, if isthe angle which a line makes with the positive direction of x-axis, then m is given by tan .

You have also learnt that the slope m of a line, passing through two points 1 1x , y and 2 2x , y is

given by 2 1

2 1

y ymx x

In this lesson, we shall find the equations of tangents and normals to different curves, using theknowledge of differential calculus. We shall also study about Rolle's Theorem and Mean ValueTheorems and their applications.


define tangent and normal to a curve (graph of a function) at a point;find equations of tangents and normals to a curve under given conditions;state Rolle's Theorem and Lagrange's Mean Value Theorem; andtest the validity of the above theorems and apply them to solve problems.

EXPECTED BACKGROUND KNOWLEDGEKnowledge of coordinate Geometry

Concept of tangent and normal to a curve

Concept of diferential coefficient of various functions

Geometrical meaning of derivative of a function at a point

MATHEMATICS

Notes

MODULE - VCalculus

300


24.1 SLOPE OF TANGENT AND NORMAL

Let y f x be a continuous curve and let

1 1P x , y be a point on it then the slope PT at

1 1P x , y is given by

1 1dy

at x , ydx ....(i)

and (i) is equal to tan

We know that a normal to a curve is a lineperpendicular to the tangent at the point of contact

We know that2

(From Fig. 24.1)

tan tan cot2

1tan

Slope of normal1 1

dymdx

at 1 1x , y or dxdy

at 1 1x , y

Note

1. The tangent to a curve at any point will be parallel to x-axis if 0 , i.e, the derivativeat the point will be zero.

i.e.dxdy at 1 1x , y 0

2. The tangent at a point to the curve y f x will be parallel to y-axis if dy 0dx

at that

point.

Let us consider some examples :

Example 24.1 Find the slope of tangent and normal to the curve2 3 2x x 3xy y 5 at (1, 1)

Solution : The equation of the curve is

2 3 2x x 3xy y 5 .....(i)

Differentialing (i),w.r.t. x, we get

Fig. 24.1

MATHEMATICS 301

Notes

MODULE - VCalculus


2 dy dy2x 3x 3 x y.1 2y 0

dx dx .....(ii)

Substituting x 1, y 1, in (ii), we get

dy dy2 1 3 1 3 1 2 0

dx dx

ordy5 8dx

dy 8dx 5

The slope of tangent to the curve at (1, 1) is 85

The slope of normal to the curve at (1, 1) is 58

Example 24.2 Show that the tangents to the curve 5 31y 3x 2x 3x6

at the points x 3 are parallel.

Solution : The equation of the curve is 5 33x 2x 3xy

6.....(i)

Differentiating (i) w.r.t. x, we get4 215x 6x 3dy

dx 6

4 2

x 3

15 3 6 3 3dy atdx 6

1 15 9 9 54 36

3 405 17 2116

4 2dy 1atx 3 15 3 6 3 3

dx 6 = 2111

The tangents to the curve at x 3 are parallel as the slopes at x 3 are equal.

Example 24.3 The slope of the curve 3 26y px q at 2, 2 is 16

.

Find the values of p and q.Solution : The equation of the curve is

3 26y px q .....(i)

MATHEMATICS

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302

Tangents and NormalsDifferentiating (i) w.r.t. x, we get

2 dy18y 2pxdx

.....(ii)

Putting x 2, y 2, we get

2 dy18 2 2p 2 4pdx

dy pdx 18

It is given equal to16

1 p p 36 18

The equation of curve becomes

3 26y 3x q

Also, the point (2, –2) lies on the curve

3 26 2 3 2 q

48 12 q or q 60 The value of p = 3, q 60

CHECK YOUR PROGRESS 24.11. Find the slopes of tangents and normals to each of the curves at the given points :

(i) 3y x 2x at x 2 (ii) 2 2x 3y y 5 at ( 1, 1 )

(iii) x a sin , y a 1 cos at 2

2. Find the values of p and q if the slope of the tangent to the curve xy px qy 2 at(1, 1) is 2.

3. Find the points on the curve 2 2x y 18 at which the tangents are parallel to the linex+y = 3.

4. At what points on the curve 2y x 4x 5 is the tangent perpendiculat to the line

2y x 7 0.

24.2 EQUATIONS OF TANGENT AND NORMAL TO A CURVEWe know that the equation of a line passing through a point 1 1x , y and with slope m is

MATHEMATICS 303

Notes

MODULE - VCalculus


1 1y y m x x

As discussed in the section before, the slope of tangent to the curve y = f (x) at 1 1x , y is given

bydydx at 1 1x , y and that of normal is 1 1

dx at x , ydy

Equation of tangent to the curve y = f (x) at the point 1 1x , y is

1 1x ,y1 1

dyy y x xdx

And, the equation of normal to the curve y = f(x) at the point 1 1x , y is

1 1

x , y1 1

1y y x xdydx

Note

(i) The equation of tangent to a curve is parallel to x-axis if x ,y1 1

dy 0dx

. In that case

the equation of tangent is 1y y .

(ii) In casex ,y1 1

dydx

, the tangent at 1 1x , y is parallel to y axis and its

equation is 1x x

Let us take some examples and illustrate

Example 24.4 Find the equation of the tangent and normal to the circle 2 2x y 25 at thepoint (4, 3)Solution : The equation of circle is

2 2x y 25 ....(i)

Differentialing (1), w.r.t. x, we getdy2x 2y 0dxdy xdx y

4,3

dy 4atdx 3

MATHEMATICS

Notes

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304

Tangents and Normals Equation of tangent to the circle at (4, 3) is

4y 3 x 43

or 4 x 4 3 y 3 0 or, 4x 3y 25

Also, slope of the normal

4,3

1 3dy 4dx

Equation of the normal to the circle at (4,3) is3y 3 x 44

or 4y 12 3x 12

3x 4y

Equation of the tangent to the circle at (4,3) is 4x+3y = 25

Equation of the normal to the circle at (4,3) is 3x = 4y

Example 24.5 Find the equation of the tangent and normal to the curve 2 216x 9y 144at

the point 1 1x , y where 1y 0 and 1x 2

Solution : The equation of curve is2 216x 9y 144 .....(i)

Differentiating (i), w.r.t. x we getdy32x 18y 0dx

ordy 16xdx 9y

As 1x 2 and 1 1x , y lies on the curve

2 216 2 9 y 144

2 80y9

4y 53

As 1y 04y 53

Equation of the tangent to the curve at 4

2, 53 is

MATHEMATICS 305

Notes

MODULE - VCalculus


4 5at 2,3

4 16xy 5 x 23 9y

or4 16 2 3y 5 x 23 9 4 5 or

4 8y 5 x 2 03 3 5

or43 5y 5 3 5 8 x 2 033 5y 20 8x 16 0 or 3 5y 8x 36

Also, equation of the normal to the curve at 4

2, 53 is

4at 2, 53

4 9yy 5 x 23 16x

4 9 2 5y 5 x 23 16 3

4 3 5y 5 x 23 8

3 8(y) 32 5 9 5 x 2

24y 32 5 9 5 x 18 5 or 9 5x 24y 14 5 0

Example 24.6 Find the points on the curve 2 2x y 1

9 16 at which the tangents are parallel

to x-axis.

Solution : The equation of the curve is

2 2x y 19 16

....(i)

Differentiating (i) w.r.t. x we get

2x 2y dy 09 16 dx

ordy 16xdx 9y

For tangent to be parallel to x-axis, dy 0dx

MATHEMATICS

Notes

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306


16x 09y

x = 0

Putting x = 0 in (i), we get 2y 16 y 4i

This implies that there are no real points at which the tangent to 2 2x y 1

9 16 is parallel to x-

axis.

Example 24.7 Find the equation of all lines having slope – 4 that are tangents to the curve1y

x 1

Solution :1y

x 1.....(i)

2dy 1dx (x 1)

It is given equal to – 4

21

4(x 1)

2 1x 14

1 3 1x 1 x ,2 2 2

Substituting 1x2

in (i), we get

1 1y 21 112 2

When3x , y 22

The points are 3 1

, 2 , , 22 2

The equations of tangents are

(a)3

y 2 4 x2

y 2 4x 6 or 4x y 8

(b)1

y 2 4 x2

y 2 4x 2 or 4x y 0

MATHEMATICS 307

Notes

MODULE - VCalculus


Example 24.8 Find the equation of the normal to the curve 3y x at (2 , 8 )

Solution : 3y x 2dy 3xdx

atx 2

dy 12dx

Slope of the normal 1

12 Equation of the normal is

1y 8 x 212

or 12(y 8) (x 2) 0 or x 12y 98

CHECK YOUR PROGRESS 24.21. Find the equation of the tangent and normal at the indicated points :

(i) 4 3 2y x 6x 13x 10x 5 at (0, 5)

(ii) 2y x at 1, 1

(iii) 3y x 3x 2 at the point whose x–coordinate is 3

2. Find the equation of the targent to the ellipse 2 2

2 2x y 1a b

at 1 1x , y

3. Find the equation of the tangent to the hyperbola

2 2

2 2x y 1a b

at 0 0x , y

4. Find the equation of normals to the curve

3y x 2x 6 which are parallel to the line x 14y 4 0

5. Prove that the curves 2x y and xy k cut at right angles if 28k 1

24.3 ROLLE'S THEOREMLet us now study an important theorem which reveals that between two points a and b on thegraph of y = f ( x) with equal ordinates f (a) and f (b), there exists at least one point c such thatthe tangent at [c,f(c)] is parallel to x-axis. (see Fig. 24.2).

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Fig. 24.2

24.3.1 Mathematical formulation of Rolle's Theorem

Let f be a real function defined in the closed interval [a, b] such that

(i) f is continuous in the closed interval [ a, b ]

(ii) f is differentiable in the open inteval ( a, b )

(iii) f (a) = f (b)

Then there is at least one point c in the open inteval (a, b) such that f '(c) 0

Remarks

(i) The remarks "at least one point" says that there can be more than one point c a , b such

that f ' c 0 .

(ii) The condition of continuity of f on [a, b] is essential and can not be relaxed

(iii) The condition of differentiability of f on ( a,b ) is also essential and can not be relaxed.

For example f(x) | x | , x [ 1,1] is continuous on [ 1,1] and differentiable on (–1, 1) andRolle's Theorem is valid for this

Let us take some examples

Example 24.9 Verify Rolle's for the function

f(x) x(x 1)(x 2),x [0,2]

Solution : f(x) x(x 1)(x 2)3 2x 3x 2x

(i) f (x) is a polynomial function and hence continuous in [0, 2]

(ii) f (x) is differentiable on (0,2)

(iii) Also f (0) = 0 and f (2) = 0

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f(0) f(2)

All the conditions of Rolle's theorem are satisfied.

Also, 2f '(x) 3x 6x 2

f '(c) 0 gives 23c 6c 2 0 6 36 24c6

1c 13

We see that both the values of c lie in (0, 2)

Example 24.10 Discuss the applicability of Rolle's Theorem forf(x) s inx sin2x , x 0, ....(i)

(i) is a sine function. It is continuous and differentiable on 0,

Again, we have, f 0 0 and f 0

f f 0 0 All the conditions of Rolle's theorem are satisfied

Now 2f '(c) 2 2cos c 1 cosc 0

or 24cos c cosc 2 0

1 1 32cosc8

1 338

As 33 6

7cosc 0.8758

which shows that c lies between 0 and

CHECK YOUR PROGRESS 24.3Verify Rolle's Theorem for each of the following functions :

(i)3 2x 5xf(x) 2x3 3

, x [0,3] (ii) 2f(x) x 1 on [–1, 1]

(iii) f(x) s inx cosx 1 on 0,2 (iv) 2f(x) x 1 x 2 on [ 1,2]

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24.4 LANGRANGE'S MEAN VALUE THEOREMThis theorem improves the result of Rolle's Theorem saying that it is not necessary that tangentmay be parallel to x-axis. This theorem says that the tangent is parallel to the line joining the endpoints of the curve. In other words, this theorem says that there always exists a point on thegraph, where the tangent is parallel to the line joining the end-points of the graph.

24.4.1 Mathematical Formulation of the TheoremLet f be a real valued function defined on the closed interval [a, b] such that(a) f is continuous on [a, b], and(b) f is differentiable in (a, b)(c) f (b) f (a)then there exists a point c in the open interval (a, b) such that

f(b) f(a)f '(c)b a

Remarks

When f (b) = f (a), f '(c) = 0 and the theorem reduces to Rolle's Theorem

Let us consider some examples

Example 24.11 Verify Langrange's Mean value theorem forf(x) (x 3)(x 6)(x 9) on [3, 5]

Solution : f(x) (x 3)(x 6)(x 9)

2(x 3)(x 15x 54)

or 3 2f(x) x 18x 99x 162 ...(i)

(i) is a polynomial function and hence continuous and differentiable in the given intervalHere, f(3) 0, f(5) (2)(–1)( 4) 8

f(3) f(5)

All the conditions of Mean value Theorem are satisfiedf(5) f(3) 8 0f '(c) 4

5 3 2

Now 2f '(x) 3x 36x 9923c 36c 99 4 or 23c 36c 95 0

36 1296 1140 36 12.5c6 6

= 8.08 or 3.9c 3.9 (3,5)

Langranges mean value theorem is verified

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Example 24.12 Find a point on the parabola 2y x 4 where the tangent is parallel to the

chord joining 4 ,0 and 5,1

Solution : Slope of the tangent to the given curve at any point is given by f '(x) at that point.

f '(x) 2 x 4

Slope of the chord joining 4 ,0 and 5,1 is

1 0 15 4

2 1

2 1

y ym

x x∵

According to mean value theorem

2 x 4 1 or1x 42

9x2

which lies between 4 and 5

Now 2y x 4

When29 9 1x , y 4

2 2 4

The required point is 9 1

,2 4

CHECK YOUR PROGRESS 24.41. Check the applicability of Mean Value Theorem for each of the following functions :

(i) 2f(x) 3x 4 on [2,3] (ii) f(x) l o g x on [1,2]

(iii)1f(x) xx

on [1,3] (iv) 3 2f(x) x 2x x 3 on [0,1]

2. Find a point on the parabola 2y x 3 , where the tangent is parallel to the chord

joining 3, 0 and 4,1

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The equation of tangent at 1 1x , y to the curve y f(x) is given by

1 1at x , y1 1y y f '(x) x x

The equation of normal at 1 1x , y to the curve y f(x) is given by

1 1x ,y1 1

1y y x xf ' x

The equation of tangent to a curve y f(x) at 1 1x , y and parallel to x-axis is

given by 1y y and parallel to y-axis is given by 1x x

Rolle's Theorem states : If f (x) is a function which is

(i) continuous in the closed interval [a,b]

(ii) differentiable in the open interval (a,b)

(iii) f(a) f(b)

then there exists a point c in (a,b) such that f '(c) = 0

Lagranges Mean Value Theorem states that if f (x) is a function which is

(i) continuous in the closed interval [a,b]

(ii) differentiable in the open interval (a,b)

(iii) f(b) f(a)

then there exists a point c in (a,b) such that

f(b) f(a)f '(c)b a


TERMINAL EXERCISE

1. Find the slopes of tangents and normals to each of the following curves at the indicatedpoints :

LET US SUM UP


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(i) y x at x=9 (ii) 3y x x at x = 2

(iii) x a( sin ) ,y a(1 cos ) at 2

(iv) 2y 2x c o s x at x = 0 (v) xy 6 at (1,6)

2. Find the equations of tangent and normal to the curve

3x acos , 3y asin at 4

3. Find the point on the curve 2 2x y 1

9 16 at which the tangents are parallel to y-axis.

4. Find the equation of the tangents to the curve

2y x 2x 5 , (i) which is parallel to the line 2x y 7 0(ii) which is perpendicular

to the line 5(y 3x) 12

5. Show that the tangents to the curve 3y 7x 11 at the points x = 2 and x 2 areparallel

6. Find the equation of normal at the point2 3am ,am to the curve 2 3ay x

7. Verify Rolle's Theorem for each of the following functions:

(i) 2f x x 1 x 2 on 1, 2 (ii)x x 2

f xx 1

on 0 ,2

(iii)28xf x 2x

3,

3x 0,

4

8. If Rolle's theorem holds for 3 2f x x bx ax, x 1, 3 with1c 23 , find the

values of a and b.

9. Verify Mean Value Theorem for each of the following functions.

(i) 2 2f x ax bx cx d on 0,1 (ii)1f x

4x 1 on 1, 4

(iii) 2y x 3 on 4,3

10. Find a point on the parabola 2f x x 3 , where the tangent is parallel to the chord

joining the points 3, 0 and 4,1

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ANSWERS


1. (i) 10, 1

10 (ii)25

, 52

(iii) 1, 1

2. p 5, q 4 3. ( 3, 3 ), ( 3, 3 ) 4. ( 3, 2 )

CHECK YOUR PROGRESS 24.21. Tangent Normal

(i) y 10x 5 x 10y 50 0

(ii) 2x y 1 x 2y 3 0

(iii)24x y 52 x 24y 483

2. 1 12 2

xx yy 1a b

3. 0 02 2

xx yy 1a b

4. x 14y 254 0, x 14y 86 0


(i)5 7c

3(ii) c 0 (iii)c

4(iv)

2 7c3


1. (i) c 2.5 (ii) 2ec 1/log (iii)c 3 (iv)

1c3

2.43 1,14 196

TERMINAL EXRCISE

1. (i)1 , 66

(ii) 113,13

(iii)1, 1 (iv) 0, not defined (v)16,6

2. 2 2 x y a; x y 0 3. 3, 0 , 3, 0

4. (i) 2x y 5 0 (ii)12x 36y 155

6. 2 22x 3my am 2 3m 0

7.2 7c

3(ii) At no real point (iii)

3c8

8. a 11, b 6

9. (i)1c2

(ii) Not applicable (iii)1c2

10.7 1

,2 4

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25

MAXIMA AND MINIMA

You are aware that in any transaction the total amount paid increases with the number of itemspurchased. Consider a function as f x 2x 1, x 0 . Let the function f(x) represent theamount required for purchasing 'x' number of items.

The graph of the function y f x 2x 1, x 0 for different values of x is shown inFig. 25.1.

Fig. 25.1

A look at the graph of the above function prompts us to believe that the function is increasing forpositive values of x i.e., for x>0 Can you think of another example in which value of the functiondecreases when x increases ? Any such relation would be relationship between time andmanpower/person(s) involved. You have learnt that they are inversely propotional . In otherwords we can say that time required to complete a certain work increases when number ofpersons (manpower) invovled decreases and vice-versa. Consider such similar function as

2g xx

, x > 0

The values of the function g (x) for different values of x are plotted in Fig. 25.2.

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Maxima and Minima

Fig. 25.2

All these examples, despite the diversity of the variables involved, have one thing in common :function is either increasing or decreasing.In this lesson, we will discuss such functions and their characteristics. We will also find outintervals in which a given function is increasing/decreasing and its application to problems onmaxima and minima.


define monotonic (increasing and decreasing) functions;

establish that dy 0dx

in an interval for an increasing function and dy 0dx

for a decreasing

function;define the points of maximum and minimum values as well as local maxima and localminima of a function from the graph;establish the working rule for finding the maxima and minima of a function using the firstand the second derivatives of the function; andwork out simple problems on maxima and minima.

EXPECTED BACKGROUND KNOWLEDGEConcept of function and types of functionDifferentiation of a functionSolutions of equations and the inquations

25.1 INCREASING AND DECREASING FUNCTIONSYou have already seen the common trends of an increasing or a decreasing function. Here wewill try to establish the condition for a function to be an increasing or a decreasing.

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Let a function f x be defined over the closed interval [a,b].

Let 1 2x , x a,b , then the function f(x) is said to be an increasing function in the given interval

if 2 1f x f x whenever 2 1x x . It is said to be strictly increasing if 2 1f x f x for all

2 1x x , 1 2x , x a,b .

In Fig. 25.3, sin x increases from -1 to +1 as x increases from 2

to 2

.

Fig. 25. 3

Note : A function is said to be an increasing function in an interval if f (x + h)> f(x) for all x belonging to the interval when h is positive.A function f(x) defined over the closed interval [a, b] is said to be a decreasing function in thegiven interval, if 2 1f x f x , whenever 2 1x x , 1 2x , x a,b . It is said to be strictly

decreasing if 1 2f x f x for all 2 1x x , 1 2x , x a,b .

In Fig. 25.4, cos x decreases from 1 t o 1 as x increases from 0 to .

Fig. 25.4

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Maxima and Minima

Note : A function is said to be a decreasing in an internal if f x h f x for all xbelonging to the interval when h is positive.

25.2 MONOTONIC FUNCTIONS

Let 1 2x , x be any two points such that 1 2x x in the interval of definition of a function f(x).Then a function f(x) is said to be monotonic if it is either increasing or decreasing. It is said to bemonotonically increasing if 2 1f x f x for all 2 1x x belonging to the interval and

monotonically decreasing if 1 2f x f x .

Example 25.1 Prove that the function f x 4x 7 is monotonic for all values of x R.

Solution : Consider two values of x say 1 2x , x R

such that 2 1x x .....(1)

Multiplying both sides of (1) by 4, we have 2 14x 4x .....(2)

Adding 7 to both sides of (2), to get

2 14x 7 4x 7

We have 2 1f x f x

Thus, we find 2 1f x f x whenever 2 1x x .

Hence the given function f x 4x 7 is monotonic function. (monotonically increasing).

Example 25.2 Show that2f x x

is a strictly decreasing function for all x < 0.

Solution : Consider any two values of x say 1 2x , x such that

2 1x x , 1 2x , x 0 .......(i)

Order of the inequality reverses when it is multiplied by a negative number. Now multiplying (i)by 2x , we have

2 2 1 2x x x x

or, 22 1 2x x x ......(ii)

Now multiplying (i) by 1x , we have

1 2 1 1x x x x

or, 21 2 1x x x .....(iii)

From (ii) and (iii), we have

2 22 1 2 1x x x x

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or, 2 22 1x x

or, 2 1f x f x .....(iv)

Thus, from (i) and (iv), we have for

2 1x x ,

2 1f x f x

Hence, the given function is strictly decreasing for all x 0.


1. (a) Prove that the functionf x 3x 4

is monotonic increasing function for all values of x R .(b) Show that the function

f x 7 2xis monotonically decreasing function for all values of x R .

(c) Prove that f x ax b where a, b are constants and a 0 is a strictly increasingfunction for all real values of x.

2. (a) Show that 2f x x is a strictly increasing function for all real x > 0.

(b) Prove that the function 2f x x 4 is monotonically increasing for

x > 2 and monotonically decreasing for 2 x 2 where x R .

Theorem 1 : If f(x) is an increasing function on an open interval ]a, b[, then its derivative

f '(x) is positive at this point for all x [a,b].

Proof : Let (x, y) or [x, f (x)] be a point on the curve y = f (x)

For a positive x, we have

x x xNow, function f(x) is an increasing function

f x x f xor, f x x f x 0

or,f x x f x 0

xx 0∵

Taking x as a small positive number and proceeding to limit, when x 0

lim f x x f xx 0 0x

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or, f ' x 0

Thus, if y f x is an increasing function at a point, then f '(x) is positive at that point.

Theorem 2 : If f (x) is a decreasing function on an open interval ]a, b[ then its derivative f'(x) is negative at that point for all x a,b .

Proof : Let (x, y) or [x, f(x)] be a point on the curve y = f(x)

For a positive x , we have x x xSince the function is a decreasing function

f x x f x x 0

or, f x x f x 0

Dividing by x , we have f x x f x 0x

x 0

or,x 0

f x x f xlim 0x

or, f ' x 0

Thus, if y f x is a decreasing function at a point, then, f ' x is negative at that point.

Note : If f x is derivable in the closed interval [a,b], then f (x) is

(i) increasing over [a,b], if f '(x)>0 in the open interval ]a,b[(ii) decreasing over [a,b], if f '(x)<0 in the open interval ]a,b[.

25.3 RELATION BETWEEN THE SIGN OF THE DERIVATIVE ANDMONOTONICITY OF FUNCTION

Consider a function whose curve is shown in the Fig. 25.5

Fig. 25.5

We divide, our study of relation between sign of derivative of a function and its increasing or

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Maxima and Minimadecreasing nature (monotonicity) into various parts as per Fig. 25.5

(i) P to R (ii) R to T (iii) T to V(i) We observe that the ordinate (y-coordinate) for every succeeding point of the curve from

P to R increases as also its x-coordinate. If 2 2x , y are the coordinates of a point that

succeeds 1 1x , y then 2 1x x yields 2 1y y or 2 1f x f x .

Also the tangent at every point of the curve between P and R makes acute angle withthe positive direction of x-axis and thus the slope of the tangent at such points of thecurve (except at R) is positive. At R where the ordinate is maximum the tangent isparallel to x-axis, as a result the slope of the tangent at R is zero.We conclude for this part of the curve that(a) The function is monotonically increasing from P to R.(b) The tangent at every point (except at R) makes an acute angle with positive

direction of x-axis.

(c) The slope of tangent is positive i.e. dy 0dx

for all points of the curve for which y is

increasing.

(d) The slope of tangent at R is zero i.e. dy 0dx

where y is maximum.

(ii) The ordinate for every point between R and T of the curve decreases though its x-

coordinate increases. Thus, for any point 2 1 2 1x x yelds y y , or 2 1f x f x .

Also the tangent at every point succeeding R along the curve makes obtuse angle withpositive direction of x-axis. Consequently, the slope of the tangent is negative for all suchpoints whose ordinate is decreasing. At T the ordinate attains minimum value and thetangent is parallel to x-axis and as a result the slope of the tangent at T is zero.We now conclude :(a) The function is monotonically decreasing from Rto T.(b) The tangent at every point, except at T, makes obtuse angle with positive direction

of x-axis.

(c) The slope of the tangent is negative i.e.,dy 0dx

for all points of the curve for which

y is decreasing.

(d) The slope of the tangent at T is zero i.e. dy 0dx

where the ordinate is minimum.

(iii) Again, for every point from T to VThe ordinate is constantly increasing, the tangent at every point of the curve between Tand V makes acute angle with positive direction of x-axis. As a result of which the slopeof the tangent at each of such points of the curve is positive.Conclusively,

dy 0dx

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Maxima and Minima

at all such points of the curve except at Tand V, where dy 0dx

. The derivativedy 0dx

on

one side, dy 0dx

on the other side of points R, T and V of the curve where dy 0dx

.

Example 25.3 Find for what values of x, the function2f x x 6x 8

is increasing and for what values of x it is decreasing.

Solution : 2f x x 6x 8

f ' x 2x 6

For f x to be increasing, f ' x 0

i.e., 2x 6 0 or, 2 x 3 0

or, x 3 0 or, x 3The function increases for x>3.

For f x to be decreasing,

f ' x 0

i.e., 2x 6 0 or, x 3 0or, x 3Thus, the function decreases for x < 3.

Example 25.4 Find the interval in which 3 2f x 2x 3x 12x 6 is increasing ordecreasing.

Solution : 3 2f x 2x 3x 12x 6

2f ' x 6x 6x 12

26 x x 2

6 x 2 x 1

For f x to be increasing function of x,

f ' x 0

i.e. 6 x 2 x 1 0 or, x 2 x 1 0

Since the product of two factors is positive, this implies either both are positive or both arenegative.

Either x 2 0 and x 1 0 or x 2 0 and x 1 0

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i.e. x 2 and x 1 i.e. x 2 and x 1

x 2 implies x 1 x 1 implies x 2

x > 2 x <–1Hence f (x) is increasing for x > 2 or x <–1.Now, for f (x) to be decreasing,

f'(x)<0or, 6 (x–2) (x+1)<0 or, (x–2) (x+1)<0Since the product of two factors is negative, only one of them can be negative, the other positive.Therefore,Either or

x – 2 > 0 and x + 1< 0 x – 2 < 0 and x + 1 > 0i.e. x > 2 and x < – 1 i.e. x < 2 and x > – 1

There is no such possibility This can be put in this formthat x > 2 and at the same timex < – 1 – 1 < x < 2

The function is decreasing in –1< x < 2.

Example 25.5 Determine the intervals for which the function

2xf x

x 1 is increasing or decreasing.

Solution :2 2

22

dx dx 1 x x 1dx dxf ' x

x 1

2

22

x 1 x 2x

x 1

2

22

1 x

x 1

22

1 x 1 xf ' x

x 1

As22x 1 is positive for all real x.

Therefore, if –1 x 0, 1 x is positive and 1 x is positive, so f ' x 0;

If 0 x 1, 1 x is positive and 1 x is positive, so f' f ' x 0;

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Maxima and Minima

If x 1, 1 x is positive and 1 x is negative, so f ' x 0;

x 1, 1 x is negative and 1 x is positive, so f ' x 0;

Thus we conclude that

the function is increasing for – 1 < x < 0 and 0 < x < 1

or, for –1 < x < 1

and the function is decreasing for x < – 1 or x > 1

Note : Points where f ' (x) = 0 are critical points. Here, critical points are x 1 , x = 1.


(a) f x c o s x is decreasing in the interval 0 x .

(b) f x x c o s x is increasing for all x.

Solution :(a) f x c o s x

f ' x s i n x

f x is decreasing

If f ' x 0i.e., s inx 0i.e., s inx 0sin x is positive in the first quadrant and in the second quadrant, therefore, sin x is positive in0 x

f x is decreasing in 0 x

(b) f x x c o s x

f ' x 1 s i n x

1 s i n xNow, we know that the minimum value of sinx is –1 and its maximum; value is 1 i.e.,sin x liesbetween –1 and 1 for all x,i.e., 1 sin x 1 or 1 1 1 sinx 1 1or 0 1 sinx 2

or 0 f ' x 2

or 0 f ' x

or f ' x 0

f x x c o s x is increasing for all x.

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CHECK YOUR PROGRESS 25.2Find the intervals for which the followiong functions are increasing or decreasing.

1. (a) 2f x x 7x 10 (b) 2f x 3x 15x 10

2. (a) 3 2f x x 6x 36x 7 (b) 3 2f x x 9x 24x 12

3. (a) 2y 3x 12x 8 (b) 2 3f x 1 12x 9x 2x

4. (a)x 2y , x 1x 1

(b)2xy , x 1

x 1(c)

x 2y , x 02 x

5. (a) Prove that the function log sin x is decreasing in ,2

(b) Prove that the function cos x is increasing in the interval , 2

(c) Find the intervals in which the function cos 2x , 0 x4

is decreasing or

increasing.Find also the points on the graph of the function at which the tangents are parallel to x-axis.

25.4 MAXIMUM AND MINIMUM VALUES OF A FUNCTIONWe have seen the graph of a continuous function. It increases and decreases alternatively. If thevalue of a continious function increases upto a certain point then begins to decrease, then thispoint is called point of maximum and corresponding value at that point is called maximum valueof the function. A stage comes when it again changes from decreasing to increasing . If the valueof a continuous function decreases to a certain point and then begins to increase, then value atthat point is called minimum value of the function and the point is called point of minimum.

Fig. 25.6

Fig. 25.6 shows that a function may have more than one maximum or minimum values. So, forcontinuous function we have maximum (minimum) value in an interval and these values are notabsolute maximum (minimum) of the function. For this reason, we sometimes call them as localmaxima or local minima.

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Maxima and MinimaA function f (x) is said to have a maximum or a local maximum at the point x = a where

a – b < a < a + b (See Fig. 25.7), if f a f a b for all sufficiently small positive b.

Fig. 25.7 Fig. 25.8

A maximum (or local maximum) value of a function is the one which is greater than all othervalues on either side of the point in the immediate neighbourhood of the point.

A function f (x) is said to have a minimum (or local minimum ) at the point x = a if

f a f a b where a – b< a < a + b

for all sufficiently small positive b.

In Fig. 25.8, the function f(x) has local minimum at the point x = a.

A minimum ( or local miunimum) value of a function is the one which is less than all other values,on either side of the point in the immediate neighbourhood of the point.

Note : A neighbourhood of a point x R is defined by open internal ]x [, when >0.

25.5 CONDITIONS FOR MAXIMUM OR MINIMUMWe know that derivative of a function is positive when the function is increasing and the derivativeis negative when the function is decreasing. We shall apply this result to find the condition formaximum or a function to have a minimum. Refer to Fig. 25.6, points B,D, F are points ofmaxima and points A,C,E are points of minima.

Now, on the left of B, the function is increasing and so f ' x 0 , but on the right of B, thefunction is decreasing and, therefore, f ' x 0 . This can be achieved only when f '(x) becomeszero somewhere in betwen. We can rewrite this as follows :A function f (x) has a maximum value at a point if (i) f '(x) = 0 and (ii) f '(x) changes sign frompositive to negative in the neighbourhood of the point at which f '(x)=0 (points taken from left toright).Now, on the left of C (See Fig. 25.6), function is decreasing and f '(x) therefore, is negative andon the right of C, f (x) is increasing and so f '(x) is positive. Once again f '(x) will be zero beforehaving positive values. We rewrite this as follows :A function f (x) has a minimum value at a point if (i) f '(x)=0, and (ii) f '(x) changes sign fromnegative to positive in the neighbourhood of the point at which f '(x) = 0.

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Maxima and MinimaWe should note here that f '(x) = 0 is necessary condition and isnot a sufficient condition for maxima or minima to exist. We canhave a function which is increasing, then constant and then againincreasing function. In this case, f '(x) does not change sign.Thevalue for which f '(x) = 0 is not a point of maxima or minima.Such point is called point of inflexion.

For example, for the function 3f x x ,x 0 is the point of

inflexion as 2f ' x 3x does not change sign as x passesthrough 0. f'(x) is positive on both sides of the value '0' (tangentsmake acute angles with x-axis) (See Fig. 25.9).

Hence 3f x x has a point of inflexion at x = 0.The points where f '(x) = 0 are called stationary points as the rate ofchange of the function is zerothere. Thus points of maxima and minima are stationary points.

RemarksThe stationary points at which the function attains either local maximum or local minimumvalues are also called extreme points and both local maximum and local minimum values arecalled extreme values of f (x). Thus a function attains an extreme value at x=a if f (a) is eithera local maximum or a local minimum.

25.6 METHOD OF FINDING MAXIMA OR MINIMAWe have arrived at the method of finding the maxima or minima of a function. It is as follows :(i) Find f'(x)(ii) Put f '(x)=0 and find stationary points(iii) Consider the sign of f '(x) in the neighbourhood of stationary points. If it changes sign from

+ve to –ve, then f (x) has maximum value at that point and if f '(x) changes sign from –veto +ve, then f (x) has minimum value at that point.

(iv) If f '(x) does not change sign in the neighbourhood of a point then it is a point of inflexion.

Example 25.7 Find the maximum (local maximum) and minimum (local minimum) points of

the function 3 2f x x 3x 9x .

Solution : Here 3 2f x x 3x 9x

2f ' x 3x 6x 9

Step I. Now 2f ' x 0 gives us 3x 6x 9 0

or 2x 2x 3 0

or x 3 x 1 0

or x 3, 1

Stationary points are x 3,x 1

Fig. 25.9

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Maxima and Minima

Step II. At x 3

For x 3 f '(x) < 0

and for x 3 f ' (x) > 0

f '(x) changes sign from –ve to +ve in the neighbourhood of 3.

f (x) has minimum value at x = 3.

Step III. At x = –1,

For x< –1, f ' (x) > 0

and for x > –1, f '(x) <0

f '(x) changes sign from +ve to –ve in the neighbourhood of –1.f(x) has maximum value at x=–1.x = –1 and x = 3 give us points of maxima and minima respectively. If we want to find

maximum value (minimum value), then we have

maximum value 3 2f 1 1 3 1 9 1

1 3 9 5

and minimum value 23f 3 3 3 3 9 3 27

1,5 and 3, 27 are points of local maxima and local minima respectively..

Example 25.8 Find the local maximum and the local minimum of the function2f x x 4x

Solution : 2f x x 4x

f ' x 2x 4 2 x 2

Putting f ' (x)=0 yields 2x–4=0, i.e., x=2.

We have to examine whether x = 2 is the point of local maximum or local minimum or neithermaximum nor minimum.

Let us take x = 1.9 which is to the left of 2 and x = 2.1 which is to the right of 2 and find f'(x) atthese points.

f ' 1.9 2 1.9 2 0

f ' 2.1 2 2.1 2 0

Since f ' x 0 as we approach 2 from the left and f ' x 0 as we approach 2 from theright, therefore, there is a local minimum at x = 2.

We can put our findings for sign of derivatives of f(x) in any tabular form including the one givenbelow :

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Example 25.9 Find all local maxima and local minima of the function

3 2f x 2x 3x 12x 8

Solution : 3 2f x 2x 3x 12x 8

2f ' x 6x 6x 12

26 x x 2

f ' x 6 x 1 x 2

Now solving f '(x)=0 for x, we get6 (x+1) (x–2) = 0

x 1, 2Thus f '(x) 0 at x = –1,2.

We examine whether these points are points of local maximum or local minimum or neither ofthem.

Consider the point x = – 1Let us take x = –1.1 which is to the left of –1 and x = – 0.9 which is to the right of 1 and findf ' (x) at these points.

f ' 1.1 6 1.1 1 1.1 2 , which is positive i.e. f '(x)>0

f ' 0.9 6 0.9 1 0.9 2 , which is negative i.e. f '(x)<0

Thus, at x = – 1, there is a local maximum.Consider the point x =2.

Now, let us take x = 1.9 which is to the left of x = 2 and x = 2.1 which is to the right of x = 2 andfind f '(x) at these points.

f ' 1.9 6 1.9 1 1.9 2= 6×(Positive number)×(negative number)= a negative number

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Maxima and Minimai.e. f '(1.9) < 0and f '(2.1)=6(2.1+1)(2.1-2), which is positivei.e., f'(2.1)>0

∵ f '(x) < 0 as we approach 2 from the leftand f ' (x) >0 as we approach 2 from the right.

x =2 is the point of local minimumThus f (x) has local maximum at x = –1, maximum value of f (x)=–2–3+12+8 =15f (x) has local minimum at x =2, minimum value of f (x)=2(8)–3(4)–12(2)+8=–12

Example 25.10 Find local maximum and local minimum, if any, of the following function

2xf x

1 x

Solution : 2xf x

1 x

Then2

22

1 x 1 2x xf ' x

1 x

2

22

1 x

1 x

For finding points of local maximum or local minimum, equate f '(x) to 0.

i.e.

2

22

1 x 01 x

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21 x 0

or 1 x 1 x 0 or x = 1,–1.Consider the value x = 1.The sign of f '(x) for values of x slightly less than 1 and slightly greater than 1 changes frompositive to negative. Therefore there is a local maximum at x =1, and the local maximum

value 21 1 1

1 1 21 1Now consider x = – 1.f '(x) changes sign from negative to positive as x passes through –1, therefore, f (x) has a localminimum at x = – 1

Thus, the local minimum value 1

2 Example 25.11 Find the local maximum and local minimum, if any, for the function

f(x) sinx cosx,0 x2

Solution : We have f(x) sinx cosxf '(x) cosx sinx

For local maxima/minima, f '(x) = 0cosx sinx 0

or, tanx 1 or, x4

in 0 x2

At x4

,

For x4

, cosx sinx

f ' x cosx sinx 0

For x4

, cosx sinx 0

f ' x cosx sinx 0

f '(x) changes sign from positive to negative in the neighbourhood of 4

.

x4

is a point of local maxima.

Maximum value f4

1 1 22 2

Point of local maxima is , 2 .4

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Find all points of local maxima and local minima of the following functions. Also , find themaxima and minima at such points.

1. 2x 8x 12 2. 3 2x 6x 9x 15

3. 3 22x 21x 36x 20 4. 4 2x 62x 120x 9

5. 2x 1 x 2 6. 2x 1

x x 2

25.7 USE OF SECOND DERIVATIVE FOR DETERMINATIONOF MAXIMUM AND MINIMUM VALUES OF A FUNCTION

We now give below another method of finding local maximum or minimum of a function whosesecond derivative exists. Various steps used are :(i) Let the given function be denoted by f (x).(ii) Find f ' (x) and equate it to zero.(iii) Solve f ' (x)=0, let one of its real roots be x = a.(iv) Find its second derivative, f "(x). For every real value 'a' of x obtained in step (iii),

evaluate f" (a). Then iff "(a) <0 then x =a is a point of local maximum.f "(a)>0 then x = a is a point of local minimum.f "(a)=0 then we use the sign of f '(x) on the left of 'a' and on the right of 'a' to arrive at the result.

Example 25.12 Find the local minimum of the following function :3 22x 21x 36x 20

Solution : Let 3 2f x 2x 21x 36x 20

Then 2f ' x 6x 42x 36

26 x 7x 6

6 x 1 x 6

For local maximum or min imum

f ' x 0

or 6 x 1 x 6 0 x = 1, 6

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Nowdf " x f ' x

dx

2d 6 x 7x 6dx

12x 42

6(2x 7)

For x =1, f " (1) =6 (2.1 – 7)= – 30 < 0

x = 1 is a point of local maximum.

and 3 2f(1) 2(1) 21(1) 36(1) 20 3 is a local maximum.

For x = 6,

f " (6)=6 (2.6 –7)=30 > 0

x = 6 is a point of local minimum

and 3 2f(6) 2(6) 21(6) 36(6) 20 128 is a local minimum.

Example 25.13 Find local maxima and minima ( if any ) for the function

f x cos4x; 0 x2

Solution : f x cos4x

f ' x 4sin4xNow, f ' x 0 4sin4x 0

or, sin4x 0 or, 4x 0, , 2

or, x 0, ,4 2

x4

0 x2

∵

Now, f " x 16cos4x

at x , f " x 16cos4

16 1 16 0

f (x) is minimum at x4

Minimum value f cos 14

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Example 25.14 (a) Find the maximum value of 32x 24x 107 in the interval [–3,–1].(b) Find the minimum value of the above function in the interval [1,3].

Solution :Let 3f x 2x 24x 107

2f ' x 6x 24

For local maximum or minimum,

f ' x 0

i.e. 26x 24 0 x 2 , 2

Out of two points obtained on solving f '(x)=0, only–2 belong to the interval [–3,–1]. We shall,therefore, find maximum if any at x=–2 only.Now f " x 12x

f " 2 12 2 24or f " 2 0which implies the function f (x) has a maximum at x =–2.

Required maximum value 32 2 24 2 107 =139

Thus the point of maximum belonging to the given interval [–3,–1] is –2 and, the maximum valueof the function is 139.(b) Now f "(x) =12 x

f" (2) = 24 > 0, [∵ 2 lies in [1, 3]]which implies, the function f (x) shall have a minimum at x =2.

Required minimum 32(2) 24(2) 107

=75 Example 25.15 Find the maximum and minimum value of the function

f x sinx 1 cosx in 0, .

Solution : We have, f x sinx 1 cosx

f ' x cosx 1 cosx sinx sinx2 2cosx cos x sin x

2 2 2cosx cos x 1 cos x 2cos x cosx 1

For stationary points, f ' x 022cos x cosx 1 0

1 1 8 1 3 1cosx 1,4 4 2

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x ,3

Now, f 0 0

3 1 3 3f sin 1 cos 1

3 3 3 2 2 4

and f 0

f (x) has maximum value 3 3

4 at x

3and miminum value 0 at x = 0 and x .


Find local maximum and local minimum for each of the following functions using second orderderivatives.

1. 3 22x 3x 36x 10 2. 3 2x 12x 5

3. 2x 1 x 2 4. 5 4 3x 5x 5x 1

5. s inx 1 cosx , 0 x2

6. s inx cosx,0 x2

7. s in2x x, x2 2

25.8 APPLICATIONS OF MAXIMAAND MINIMA TOPRACTICAL PROBLEMS

The application of derivative is a powerful tool for solving problems that call for minimising ormaximising a function. In order to solve such problems, we follow the steps in the followingorder :(i) Frame the function in terms of variables discussed in the data.(ii) With the help of the given conditions, express the function in terms of a single variable.(iii) Lastly, apply conditions of maxima or minima as discussed earlier.

Example 25.16 Find two positive real numbers whose sum is 70 and their product ismaximum.Solution : Let one number be x. As their sum is 70, the other number is 70 x. As the twonumbers are positive, we have, x 0,70 x 0

70 x 0 x 70

0 x 70

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Maxima and MinimaLet their product be f (x)

Then 2f(x) x(70 x) 70x x

We have to maximize the prouct f (x).We, therefore, find f '(x) and put that equal to zero.

f '(x) = 70 –2xFor maximum product, f ' (x) = 0or 70 –2x = 0or x = 35Now f"(x) = –2 which is negative. Hence f (x) is maximum at x = 35The other number is 70 x 35Hence the required numbers are 35, 35.

Example 25.17 Show that among rectangles of given area, the square has the least perimeter.Solution : Let x, y be the length and breadth of the rectangle respectively.

Its area = xySince its area is given, represent it by A, so that we have

A = xy

orAyx

... (i)

Now, perimeter say P of the rectangle = 2 (x + y)

orA

P 2 xx

2dP A

2 1dx x

...(ii)

For a minimum dPP, 0.dx

i.e. 2A

2 1 0x

or 2A x or A x

Now,2

2 3d P 4Adx x

, which is positive.

Hence perimeter is minimum when x AAyx

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2x xx

2A x∵

Thus, the perimeter is minimum when rectangle is a square.

Example 25.18 An open box with a square base is to be made out of a given quantity of sheet

of area 2a . Show that the maximum volume of the box is 3a

6 3.

Solution : Let x be the side of the square base of the box and y its height.

Total surface area of othe box 2x 4xy

2 2x 4xy a or2 2a xy4x

Volume of the box, V base area height2 2

2 2 a xx y x

4x

or 2 31V a x x4

...(i)

2 2dV 1 a 3xdx 4

For maxima/minimadV 0dx2 21 a 3x 0

42

2 a ax x3 3

...(ii)


Volume3 3 3a1 a a

4 3 3 3 6 3..(iii)

Again2

2 22

d V d 1 3a 3x xdx 4 2dx

x being the length of the side, is positive.2

2d V 0dx

The volume is maximum.

Hence maximum volume of the box 3a

6 3.

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Example 25.19 Show that of all rectangles inscribed in a given circle, the square has themaximum area.Solution : Let ABCD be a rectangle inscribed in a circle of radius r. Then diameter AC= 2rLet AB = x and BC = y

Then 2 2 2AB BC AC or 22 2 2x y 2r 4r

Now area A of the rectangle = xy

2 2A x 4r x

2 22 2

x 2xdA 4r x 1dx 2 4r x

2 2

2 2

4r 2x

4r x

For maxima/minima, dA 0dx

2 2

2 2

4r 2x0 x 2r

4r x

Now

2 2 2 22 2 2

2 2 2

2x4r x 4x 4r 2x

d A 2 4r xdx 4r x

2 2 2 2

32 2 2

4x 4r x x 4r 2x

4r x

2

32 2

4 2 2r 0

2r... (Putting x 2r )

3

38 2r 4 0

2 2r

Thus, A is maximum when x 2r

Now, from (i), 2 2y 4r 2r 2 r

x = y. Hence, rectangle ABCD is a square.

Fig. 25.10

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Example 25.20 Show that the height of a closed right circular cylinder of a given volume andleast surface is equal to its diameter.Solution : Let V be the volume, r the radius and h the height of the cylinder.Then, 2V r h

or 2Vhr

... (i)

Now surface area 2S 2 rh 2 r

2 22

V 2V2 r. 2 r 2 rrr

Now 2dS 2V 4 rdr r

For minimum surface area, dS 0dr

22V 4 r 0r

or 3V 2 r

From (i) and (ii), we get3

22 rh 2r

r..(ii)

Again,2

2 3d S 4V 4 8 4dr r

... [Using (ii)]

12 0 S is least when h = 2r

Thus, height of the cylidner = diameter of the cylinder.

Example 25.21 Show that a closed right circular cylinder of given surface has maximum

volume if its height equals the diameter of its base.Solution : Let S and V denote the surface area and the volume of the closed right circularcylinder of height h and base radius r.

Then 2S 2 rh 2 r .....(i)( Here surface is a constant quantity, being given)

2V r h2

2 S 2 rV r2 r

2r S 2 r2

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Maxima and Minima

3SrV r2

2dV S 3rdr 2

For maximum or minimum, dV 0dr

i.e., 2S 3r 02

or 2S 6 r

From (i), we have 2 26 r 2 rh 2 r

24 r 2 rh

2r h ...(ii)

Also,2

22

d V d S 3 rdr 2dr

6 r,d S

0d r 2

∵

= a negative quantity Hence the volume of the right circular cylinder is maximum when its height is equal to twice itsradius i.e. when h = 2r.

Example 25.22 A square metal sheet of side 48 cm. has four equal squares removed from thecorners and the sides are then turned up so as to form an open box. Determine the size of thesquare cut so that volume of the box is maximum.Solution : Let the side of each of the small squares cut be x cm, so that each side of the box tobe made is (48–2x) cm. and height x cm.Now x > 0, 48–2x > 0, i.e. x < 24

x lies between 0 and 24 or 0 x 24Now, Volume V of the box

48 2x 48 2x x

i.e. 2V 48 2x x

2dV 48 2x 2 48 2x 2 xdx

48 2x 48 6x

Fig. 25.11

Fig. 25.12

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Condition for maximum or minimum is dV 0dx

i.e., 48 2x 48 6x 0

We have either x 24 , or x = 8

∵ 0 < x < 24 Rejecting x = 24, we have, x = 8 cm.

Now,2

2d V 24x 384dx

2

2x 8

d V 192 384 192 0,dx

Hence for x = 8, the volume is maximum.Hence the square of side 8 cm. should be cut from each corner.

Example 25.23 The profit function P (x) of a firm, selling x items per day is given by

P(x) 150–x x 1625 .

Find the number of items the firm should manufacture to get maximum profit. Find the maximumprofit.Solution : It is given that 'x' is the number of items produced and sold out by the firm every day.In order to maximize profit,

dPP ' x 0 i.e. 0dx

ord 150 x x 1625 0

dxor 150–2x = 0or x 75

Now, d P ' x P " x 2 adx

negative quantity. Hence P (x) is maximum for x = 75.

Thus, the firm should manufacture only 75 items a day to make maximum profit.

Now, Maximum Profit P 75 150 75 75 1625

= Rs. (75×75–1625)

= Rs. (5625–1625)

= Rs. 4000

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Example 25.24 Find the volume of the largest cylinder that can be inscribed in a sphere ofradius 'r' cm.Solution : Let h be the height and R the radius of the base of the inscribed cylinder. Let V be thevolume of the cylinder.

Then 2V R h ...(i)From OCB, we have

22 2hr R

2... 2 2 2OB OC BC∵

22 2 hR r

4...(ii)

Now2 3

2 2h hV r h r h4 4

22dV 3 hr

dh 4

For maxima/minima, dV 0dh

22 3 hr 0

42

2 4rh3

2rh3

Now2

2d V 3 h

2dh2

2d V 2r 3 2ra t h

3 2 3dh

3 r 0

V is maximum at 2rh3

Putting 2rh3

in (ii), we get

2 22 2 4r 2rR r

4 3 32

R r3

Maximum volume of the cylinder 2R h3

2 32 2r 4 rr cm .3 3 3 3

Fig. 25.13

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CHECK YOUR PROGRESS 25.51. Find two numbers whose sum is 15 and the square of one multiplied by the cube of the

other is maximum.2. Divide 15 into two parts such that the sum of their squares is minimum.3. Show that among the rectangles of given perimeter, the square has the greatest area.4. Prove that the perimeter of a right angled triangle of given hypotenuse is maximum when

the triangle is isosceles.5. A window is in the form of a rectangle surmounted by a semi-circle. If the perimeter be

30 m, find the dimensions so that the greatest possible amount of light may be admitted.6. Find the radius of a closed right circular cylinder of volume 100 c.c. which has the minimum

total surface area.7. A right circular cylinder is to be made so that the sum of its radius and its height is 6 m.

Find the maximum volume of the cylinder.8. Show that the height of a right circular cylinder of greatest volume that can be inscribed

in a right circular cone is one-third that of the cone.9. A conical tent of the given capacity (volume) has to be constructed. Find the ratio of the

height to the radius of the base so as to minimise the canvas requried for the tent.10. A manufacturer needs a container that is right circular cylinder with a volume 16 cubic

meters. Determine the dimensions of the container that uses the least amount of surface(sheet) material.

11. A movie theatre's management is considering reducing the price of tickets from Rs.55 inorder to get more customers. After checking out various facts they decide that the averagenumber of customers per day 'q' is given by the function where x is the amount of ticketprice reduced. Find the ticket price othat result in maximum revenue.

q = 500+100 xwhere x is the amount of ticket price reduced. Find the ticket price that result is maximumrevenue.

Increasing function : A function f (x) is said to be increasing in the closed interval [a,b]

if 2 1f x f x whenever 2 1x x

Decreasing function : A function f (x) is said to be decreasing in the closed interval[a,b]

if 2 1f x f x whenever 2 1x x

f (x) is increasing in an open interval ]a,b[ if f '(x) > 0 for all x [a,b]

f (x) is decreasing in an open interval ]a,b[

LET US SUM UP

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Maxima and Minima

if f '(x) < 0 for all x [a,b]

Monotonic function :(i) A function is said to be monotonic (increasing) if it increases in the given interval.(ii) A function f (x) is said to be monotonic (decreasing) if it decreases in the given

interval.A function f (x) which increases and decreases in a given interval, is not monotonic.In an interval around the point x = a of the function f (x),(i) if f ' (x) > 0 on the left of the point 'a' and f '(x) < 0 on the right of the point x = a,

then f (x) has a local maximum.(ii) if f '(x) <0 on the left of the point 'a' and f ' (x) >0 on the right of the point x = a,

then f (x) has a local minimum.If f (x) has a local maximum or local minimum at x = a and f (x) is derivable at x = a, thenf '(a) = 0(i) If f '(x) changes sign from positive to negative as x passes through 'a', then f (x) has

a local maximum at x = a.(ii) If f ' (x) changes sign from negative to positive as x passes othrough 'a', then f (x)

has a local minimum at x = a.Second order derivative Test :(i) If f '(a) = 0, and f "(a) < 0; then f (x) has a local maximum at x = a.(ii) If f '(a) = 0, and f "(a) >0; then f (x) has a local minimum at x = a.(iii) In case f '(a) = 0, and f " (a) = 0; then to determine maximum or minimum at x = a,

we use the method of change of sign of f '(x) as x passes through 'a' to.


TERMINAL EXERCISE

1. Show that 2f(x) x is neither increasing nor decreasing for all x R .

Find the intervals for which the following functions are increasing or decreasing :

2. 3 22x 3x 12x 6 3.x 4 , x 04 x

4. 4 2x 2x 5. sinx cosx,0 x 2


MATHEMATICS 345

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Maxima and MinimaFind the local maxima or minima of the following functions :

6. (a) 3 2x 6x 9x 7 (b) 32x 24x 107(c) 3 2x 4x 3x 2 (d) 4 2x 62x 120x 9

7. (a) 21

x 2(b)

xx 1 x 4 ,1 x 4

(c) x 1 x , x 1

8. (a)1sinx cos2x2

, 0 x2

(b) sin 2x ,0 x 2

(c) x 2sinx,0 x 29. For what value of x lying in the closed interval [0,5], the slope of the tangent to

3 2x 6x 9x 4is maximum. Also, find the point.

10. Find the vlaue of the greatest slope of a tangent to3 2x 3x 2x 27 at a point of othe curve. Find also the point.

11. A container is to be made in the shape of a right circular cylinder with total surface area of24 sq. m. Determine the dimensions of the container if the volume is to be as large aspossible.

12. A hotel complex consisting of 400 two bedroom apartments has 300 of them rented andthe rent is Rs. 360 per day. Management's research indicates that if the rent is reduced by

x rupees then the number of apartments rented q will be 5q x 300,0 x 804

.

Determine the rent that results in maximum revenue. Also find the maximum revenue.

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Maxima and Minima

ANSWERS


1. (a) Increasing for 7x2

, Decreasing for 7x2

(b) Increasing for5x2

, Decreasing for 5x2

2. (a) Increasing for x 6 or x 2 , Decreasing for 2 x 6(b) Increasing for x 4 or x 2 , Decreasing for x in the interval ]2,4[

3. (a) Increasing for x 2; decreasing for x 2(b) Increasing in the interval 1 x 2 , Decreasing for x 1 or x 2

4. (a) Increasing always.(b) Increasing for x 2 , Decreasing in the interval 0 x 2(c) Increasing for x 2 or x 2 Decreasing in the interval 2 x 2

5. (c) Increasing in the interval3 7x8 8

Decreasing in the interval30 x8

Points at which the tangents are parallel to x-axis are 3x8

and7x8

CHECK YOUR PROGRESS 25.31. Local minimum is – 4 at x = 42. Local minimum is 15 at x = 3, Local maximum is 19 at x = 1.3. Local minimum is – 128 at x = 6, Local maximum is – 3 at x = 1.4. Local minimum is – 1647 at x = – 6, Local minimum is – 316 at x = 5,

Local maximum is 68 at x = 1.5. Local minimum at x = 0 is – 4, Local maximum at x = – 2 is 0.

6. Local minimum at x = – 1, value 1 Local maximum at x = 3, value 17

CHECK YOUR PROGRESS 25.41. Local minimum is – 34 at x = 2, Local maximum is 91 at x = – 3.2. Local minimum is – 5 at x = 0, Local maximum is 251 at x = 8.3. Local minimum – 4 at x = 0, Local maximum 0 at x = – 2.4. Local minimum = – 28; x = 3, Local maximum = 0 ; x = 1.

Neither maximum nor minimum at x = 0.

MATHEMATICS 347

Notes

MODULE - VCalculus

Maxima and Minima

5. Local maximum 3 3 ; x

4 3

6. Local maximum 2 ; x4

7. Local minimum 3 ; x

2 6 6, Local maximum

3 ; x2 6 6


1. Numbers are 6,9. 2. Parts are 7.5, 7.5

5. Dimensions are : 30 30,

4 4 meters each.

6. radius1350 cm ; height

13502 cm

7. Maximum Volume 32 cubic meters. 9. h 2r10. r = 2 meters, h = 4 meters. 11. Rs. 30,00

TERMINAL EXERCISE

2. Increasing for x > 2 or < –1, Decreasing in the interval –1 <x <23. Increasing for x > 4 or x < – 4, Decreasing in the interval ]–4,4[4. Increasing for x > 1 or – 1 < x < 0, Decreasing for x < –1 or 0 < x < 1

5. Increasing for 30 x4

or 7 x 24

, Decreasing for 3 7x .4 4

6. (a) Local maximum is 11 at x =1 ; local minimum is 7 at x = 3.(b) Local maximum is 139 at x = –2; local minimum is 75 at x = 2.

(c) Local maximum is 20 at x = –3; local minimum is 4027

at 1x3

.

(d) Local maximum is 68 at x = 1; local minimum is–316 at x = 5 and– 1647 at x = – 6.

7. (a) Local minimum is 12

at x = 0. (b) Local maximum is –1 at x = 2.

(c) Local maximum is 2 2at x .

33 3

8. (a) Local maximum is 3 at x ;4 6

Local minimum is 1 at x ;2 2

MATHEMATICS

Notes

MODULE - VCalculus

348

Maxima and Minima

(b) Local maximum is 1 at 45x and x4

; Local minimum is –1 at 34x

(c) Local maximum is 4 3 at x ;3

Local minimum is 53

53 at x .3

9. Greatest slope is 24 at x = 5; Coordinates of the point: (5, 24).10. Greatest slope of a tangent is 5 at x = 1, The point is (1,–23)11. Radius of base = 2 m, Height of cylinder = 4 m.12. Rent reduced to Rs. 300, The maximum revenue = Rs. 1,12,500.

348/1

Partial Differentiation

Objectives

After studying this lesson, you will be able to know.

1. Definition of partial derivative corresponds closed to that of derivative of a function of a single

variable.

2. The existence of first order partial derivatives does not guarantee the continity of the function.

3. Discuss about the functions with two independent variables.

25.9 Partial Differentiation

So far we have studied the concepts of limits, continuity, differentiability of functions of a single

variable. In this chapter we consider these concepts for function of two or more real variables. Most

of the definitions and theorems apply to function of n independent variables as well, but we restrict our

discussion to functions of two independent variables only.

Recall that the cartesian product A × B of sets A and B is the collection of all ordered pairs

(a, b) with a ∈ A and b ∈ B. If A ⊆ R and B ⊆ R, A × B is a subset of R × R i.e., R2. The

elements of R2 can be considered to be points in the two dimensional plane. If fact, if (a, b) ∈ R2 , it

can be identified with a unique point with cartesian coordinates (a, b) in the plane.

Suppose A ⊆ R, B ⊆ R and E = A × B so that E ⊆ R2. Then a function f : E → R is called a

real valued function of two real variables. Also for (x, y) ∈ E its image under f is denoted by

348/2

f(x, y) is a real number, say Z, for any (x, y) ∈ E, So that we can write z = f(x, y). Some times it is

convenient to write z = z(x, y), where the function it self is denoted by z. Throughout this chapter by

a function of two variables we mean a real valued function of two real variables.

25.9.1 Partial Derivatives

In in the function f(x,y), y is held constant and the resulting function of x is differentiated with

respect to x, then the resulting derivative is called the partial derivative of f with respect to x, That

is denoted by f

x

∂∂ or fx

That is, we define 0

( , ) ( , )limxh

f f x h y f x yf

x h→

∂ + −= =∂ provided the limit exists.

Similarly, the partial dervative of f with respect to y denoted by f

y

∂∂

or fy is defined as

0

( ) ( , )limyh

f f x y h f x yf

y h→

∂ + + −= =∂ provided the limit exists. The partial derivatives

,f f

x y

∂ ∂∂ ∂

are called first order partial derivative of f with respect to x and y respectively.

Example 1

If f(x, y) = x2y2 − y sin x, then

2 (2 ) cosf

y x y xx

∂ = −∂

= 2 xy2 − y cos x

2 2(2 ) sin 2 sin .f

x y x x y xy

∂ = − = −∂

One can notice that the definition of a partial derivative corresponds closely to that of an ordinary

derivative of a function of a single variable. Consequently many of the rules of differentiation of function

of a single variable remain valid in the case of partial differentiation. However, the existance of all the

first order partial derivatives at a point does not guarantee the continuity of the function there. The

following example illustrates if.

348/3

Example 2.

Let f(x, y) = 2 2

x y

x y+, if (x, y) ≠ (0, 0). If (x, y) = (0, 0) i.e., f(0, 0) = 0 For h ≠ 0, we have

( ,0) (0,0)0

f h f

h

− =

Hence 0

( ,0) (0,0)limh

f h f

h→

− exists and is equal to zero.

Hence fx|(0, 0) = 0

Similarly fy|(0, 0) = 0

We have already noted that f(x, y) is not continuous at (0, 0_

Thus the existance of partial derivatives may not guarantee the conitnuity of f.

25.9.2 The Second order partial derivatives

As the partial derivatives of a function f(x, y) are themselves function of x and y , they inteurn

may process partial derivatives. We define these second order partial derivatives as follows :

( )2

2 xf f

fx x xx

∂ ∂ ∂ ∂ = = ∂ ∂ ∂∂

0

( , ) ( , )lim x x

h

f x h y f x y

h→

+ −=

and ( )2

2 yf f

fy y yy

∂ ∂ ∂ ∂= = ∂ ∂ ∂∂

0

( , ) ( , )lim y y

k

f x y k f x y

k→

+ −=

provided these limits exists.

In addition to these, There are two other mixed second order partial derivatives, namely

( )2

xf f

fy x y x y

∂ ∂ ∂ ∂ = = ∂ ∂ ∂ ∂ ∂

348/4

0

( , ) ( , )lim x x

k

f x y k f x y

k→

+ −=

and ( )2

yf f

fx y x y x

∂ ∂ ∂ ∂= = ∂ ∂ ∂ ∂ ∂

0

( , ) ( , )lim y y

h

f x h y f x y

h→

+ −=

Note that2

xyf

fy x

∂ =∂ ∂ for

( )2

[ ]x x y xyf f

f f fy x y x y

∂ ∂ ∂ ∂ = = = = ∂ ∂ ∂ ∂ ∂

similarly, 2

yxf

fx y

∂ =∂ ∂

Note also that2 2

2 2,xx yy

f ff f

x y

∂ ∂= =∂ ∂

we shall confine to those functions for which fxy = fyx.

Example 3

Find all the first and second order partial derivatives for f(x, y) = e2x−y .

Sol: For the given function

fx = 2e2x−y , fxx = 4 e2x−y

fy = − e2x−y , fyy = e2x−y

fxy = −2 e2x−y , fxx = − 2 e2x−y

Example 4

Find all the first and second order partial derivatives for f(x, y) = cos(ax + by) where a and b

are real constants.

Sol : For the given function

fx = − a sin(ax + by) , fy = − b sin (ax + by)

fxx = − a2 cos(ax + by) , fyy = − b2 cos (ax + by)

348/5

fxy = [fx]y = − ab cos (ax + by)

fyx = [fy]x = −ab cos (ax + by)

Example5

If z = eax sin by where a and b are real constants find Zx. Zy, Zxx, Zyy, Zyx .

Sol :

Given Z = eax sin by,

Zx = a eax sin by, Zy = b eax cos by

Zxx = (Zx)x = a2 eax sin by, Zyy = (Zy)y = − b2 eax sin by

Zxy = (Zx)y = ab eax cos by and

Zyx = (Zy)x = ab eax cos by

Example 6

If f = x2 − y2, show that fxx + fyy = 0.

Sol: Given that 2 2x xxf x f= ⇒ =

2 2y yyf y f= − ⇒ = −

2 2 0xx yyf f+ = − =

∴ 0xx yyf f+ =

25.10 Homogeneous Functions

We shall discuss an important class of functions, namely the class of homogeneous functions and

an important theorem due to Euler, related to such functions. First let us give a formal definition of a

homogeneous function.

25.10.1 Defintion

A non-zero function f = f(x, y) is said to be homogeneous if there is a real number α such that

f (λx, λy) = λα f(x, y) for all those values of k for which both sides of the above equation are

meaningful. The number α is called the degree of the homogeneous function f .

348/6

Example 7 :

f(x, y) = x3 + y3 is a homogeneous function of degree 3, why becuase

f(λx, λy) = λ3[x3 + y3]

= λ3 f(x, y)

Example 8 :

f(x, y) = (x2 + 4y2)−2/7, is a homogeneous function of degree4

7

−. Since

2 2 2 2 2 / 7( , ) ( 4 )f x y x y −λ λ = λ + λ

4 / 7( , ) ( , )f x y f x y−λ λ = λ


1. Find ,z z

x y

∂ ∂∂ ∂

for given functions.

(i) cos

cosec

xz

y= (ii) z = 3y + x ey

(iii) 1Tanx

zy

− =

(iv) 2

logy

z xx

= +

(v) z = y ey + x ex (vi) z= sin (y2 − x)

2. If z = ex cos y then find 2 2

2 2,

z z

x y

∂ ∂∂ ∂

3. If 2 2

yf

x y=

+ show that fxx + fyy = 0.

4. If z = ex+y + f(x) + g(y) then show that zxy = ex+y.

5. If z = log (Tan x + Tan y). Then show that (sin 2x) zx + (sin 2y) zy = 2.

348/7

6. If u(x, y) = 2

3 3

x y

x y+ Then show that x ux + y uy = 0.

Let Us Sum Up

1.0

( , ) ( , )limxh

f f x h y f x yf

x h→

∂ + −= =∂

0

( , ) ( , )limyh

f f x y k f x yf

y k→

∂ + −= =∂

2.2

2 0

( , ) ( , )( ) lim x x

xh

f x h y f x yf ff

x x y hx →

+ −∂ ∂ ∂ ∂ = = = ∂ ∂ ∂∂

= fxx

2

2 0

( , ) ( , )( ) lim y y

yk

f x y k f x yf ff

y y y ky →

+ − ∂ ∂ ∂ ∂= = = ∂ ∂ ∂∂

= fyy

2

0

( , ) ( , )( ) lim x x

xk

f x y k f x yf ff

y x y x y k→

+ −∂ ∂ ∂ ∂ = = = ∂ ∂ ∂ ∂ ∂

= fxy

2

0

( , ) ( , )( ) lim y y

yh

f x h y f x yf ff

x y x y x h→

+ − ∂ ∂ ∂ ∂= = = ∂ ∂ ∂ ∂ ∂

= fyx

Supportive Websites



348/8

Answers


(1) (i) zx = − sinx siny, zy = cos

(ii) zx = ey, zy = 3 + x ey

(iii)2 2 2 2

,x yy x

z zx y x y

−= =+ +

(iv) 3 3

1,

( )x y

x zyz z

x x y y x

3 −= =+ +

(v) zx = x ex + ex, zy = y ey + ey

(vi) zx = − cos(y2 − x), zy = 2y cos(y2 − x)

(2)2

2cosxz

e yx

∂ =∂

2

2cos xz

y e zy

∂ = − = −∂

348/9

Successive Differentiation

Second and higher order derivatives have applications in curve sketching and in the representa-

tion of functions as infinite series. Further, they are useful in differential equations, complex analysis

etc.,

Objectives

From this chapter we can get the following points.

� curve sketching

� Representation of functions as infinite series

� nth order derivatives of some standard functions.

25.11 Successive Differentiation - nth derivatives

In this section we define first and higher order derivatives of a real valued function f and find the

nth order derivatives of some standard functions.

Definition

Let f be a real valued function defined on a subset E of R. Suppose that E contains a

neighbour hood (left or right of both) of each of its points. We known that f is the limit of

( ) ( )f c h f c

h

+ − as h tends to 0 exists. This limit is denoted by f'(c) and is called the derivative of

f at 'c' or the differential coefficient of f at 'c'. It f is differentiable at every point x of its domain E,

348/10

then we can define a function f ' on E as f'(x) is the limit of ( ) ( )f x h f x

h

+ − as h tends to 0. The

function f ' thus defined is called the first derivative or first order derivative of f.

Again of the function f ' is differentiable at every point x of E, then we can define function f "

on E as

0

( ) ( )( )

x

f x h f xf x Lt

h→

′ ′+ −′′ =

as h tends to 0.

The function thus defined is called the second derivative or second orderivative of f. This

process can be continued as long as the successive derivatives are differentiable at every point of E. In

general, the nth derivative or nth order derivative of f is the derivative of f (n−1) and is denoted by

f (n) (i.e., f " .....(n dasehes)). Thus

f n(x)( 1) 1

0

( ) ( )n n

h

f x h f xLt

h

− −

→

+ −=

as h tends to 0.

If y = f (x) then the notations for the nth derivative f are

, D , ,n n

nnn n

d f d yf y

dx dx

which are used according to our convenience. Note that

( ) [ ( )]d

f x f xdx

′ = and( ) ( 1)( ) [ ( )]n nd

f x f xdx

−=

f n(x) = n

n

d y

dx= yn = Dny

1

1[ ( )] for 1.

n

n

df x n

dx

−

−′= >

If f (n)(x) exists for all positive integers n and for all x in E, then f is said to be an infinitely

differntiable function on E.

The above porcess of finding higher order derivative of f is called successive differentiation.

348/11

25.11.1 nth order derivatives of some standard functions

(i) Let a, b and m be real numbers. Suppose that a ≠ 0 and f(x) = (ax + b)m Then

f n(x) = m(m − 1) ... (m − n + 1) an (ax + b)m−n (n∈ N)

whenever ax + b > 0 and n is a positive integer.

Note: In the above theorem if m is a positive integer, then m = n impllies that f (n)(x) = n! an andhence f (n)(x) = 0 for n > m.

If m is a positive integer such that m > n, then

f (n)(x) = m(m−1) ..... (m−n+1) an (ax + b)m−n

!( )

( )!

nm nm a

ax bm n

−= +−

(ii) On taking m = −1 in theorem 7.1.4, we get

1 1( ) ( )f x ax b

ax b−= + =

+ ,

and f (n)(x) = (−1) (−2)..... (−n) an (ax + b)−1−n

that is, 1

1 ( 1) !.

( )

n n n

n n

d n a

ax bdx ax b +− = + +

(iii) Let f(x) = eax+bfor all x in R. Then f n(x) = aneax+bfor all x in R and for all positive integers n.

Example 1.

Find the 4th derivative of 1

2 3x+ .

Sol: 1

1 ( 1) !( ) ( )

( )

n nn

n

n af x f x

ax b ax b +−= ⇒ =

+ +

by above theorem, taking a = 2, b = 3 and m = −1 we get

f 4(x) = (−1) (−2)(−3)(−4) 24 (2x + 3)−5

4

5

4! 2

(2 3)x=

+

348/12

Example 2

Find the 5th derivative of sin 6x.

Sol: [sin ( )] sin N2

ax bn

nn

d na ax b n

dx+ π = + + ∀ ∈

The above formula taking a = 6, b = 0, n = 5

D5(sin 6x) = 65 sin 5

62

xπ +

= 65 sin 62

xπ +

= 65 cos 6x.

Example 3

Show that f(x) = x + cot x satisfies (sin2x) f"(x) + 2x = 2 f(x).

Sol : Now f'(x) = 1 − cosec2x ......... (1)

so that (sin2x) f'(x) = sin2 x − 1 ....... (2)

Differentiating both sides of (2) w.r.t x,

(sin2x) f"(x) + 2 sinx cosx f'(x) = 2 sin x cos x

Hence

sin2x f"(x) = 2 sinx cosx(1 − f'(x))

= 2 sinx cosx cosec2x (1) =Å¡

= 2 cotx = 2 [f(x) − x]


I 1. Ifa bx

yc dx

+=+ , Then show that 2 y1 y3 = 3y2

2

2. If y = a + b e−4x , Then show that y2 + 4y1 = 0 .

3. If y = axn+1 + b x−n , then show that x2 y2 = n(n + 1)y.

4. If y = x + Tanx, then show that y2 cos2x + 2x = 2y.

348/13

5. If a y4 = (x + b)5 , then show that 5y y2 = y12 .

6. If ax2 + 2h xy + by2 = 1. Then show that 2 2

2 3( )

d y h ab

dx hx by

−=+

.

II. Find the nth derivative of following functions.

1. Find the nth derivative of cos2x.

2. Find the nth derivative of 1

3 4x+ .

3. If 2

( 1)( 2)y

x x=

− − Then find yn .

25.12 Leibnitz Theorem and its Applications

Let f and g be two functions having nth order derivatives over a domain E ⊆ R. We have seen

in the previous section that f + g and f − g, kf(k being a constant) have nth order derivatives over

E and that

(f + g)(n) = f (n) + g(n)

(f − g)(n) = f (n) − g(n)

(k f)(n) = k f (n)

In this section we obtain a formula for the nth derivative of the product f g of two functions

f and g .

25.12.1 Theorem (Leibnitz Theorem)

Suppose that n is a positive integer. Let f and g be two functions defined on an interval I or R

having nth order derivative on I. The fg has nth other derivative on I and

(f g(n)(x) = f n(x) . g(x) + n c1 f (n−1)(x) g'(x) + ..... + n cr f (n−r)(x) g(r)(x) + .... + f(x) g(n)(x)

( ) ( )

0

( ) ( ) Ir

nn r r

cr

n f x g x x−

== ∀ ∈∑

Note: Where u and v are used for f(x) and g(x), nth derivative of u denoted by un and u0 = u.

0 10 1 1 0( ) .... ....r nn c n c n c n r r c nuv n u v n u v n u v n u v− −= + + + + +

348/14

Example 4

If u and v are function of x having 4th order derivatives, write the formula for D4(uv).

Sol: By Leibnitz theorem

D4(uv) = D4(u) v + 4c1 D3(u). D(v) + 4c2 D2(u) D2(v) + 4c3 D(u) D3v + u D4(v)

= u4 v + 4 u3 v1 + 6 u2 v2 + 4 u1 v3 + u v4

Example 5

Find D4 (x log x) at x = e, x > 0

Sol: By Leibnitz Theorem

D4 (x log x) = D4 (log x . x)

= D4 (log x) x + 4 c1 D3 (log x) D(x)

(Since higher derivatives of x vanish

4 3

6 2. 4. .1x

x x

−= +

3 3 3

6 8 2

x x x

−= + =

∴ at x = e

D4(x log x) = 3

2

e

Example 6

Find the nth derivative of x ex.

Sol: By Leibnitz Theorem

Dn (x ex) = Dn(ex x)

= Dn(ex) x + nc1 Dn−1(ex) D(x)

= ex . x + n ex

= (x + n) ex

348/15


I. 1. Find D3(sin x log x) when 2

xπ= .

2. Find the sum of the middle terms in the Leibnitz expansion of D3(ex cos x).

II(1) Find the nth derivative of the following using Leibnitz theorem.

(i) h(x) = e3x+5 x2

(ii) x3 eax

(iii) x2 cos2x

(2) If y = xn log x, then show that Õ x yn+1 = n! .

(3) If y = cos(m log x) (x > 0) then show that x2 y2 + xy1 + m2 y = 0 and hence deduce that

x2 yn+2 + (2n + 1) x yn+1 + (m2 + n2) yn = 0.

(4) If y = esin−1x, then show that (1 − x2) y2 − x y1 − y = 0 and hence deduce that

(1 − x2) yn+2 − (2n + 1) x yn+1 − (n2 + 1) yn = 0

Let Us Sum Up

Function f(x) nth derivative f (n)(x)

1. (ax + b)m m(m−1) .... (m−n+1) an (ax + b)m−n

2.1

ax b+ 1

( 1) !

( )

n n

n

n a

ax b +−

+

3. eax+b an eax+b

4. sin(ax+ b) an sin(ax + b +2nπ

)

5. cos (ax + b) an cos(ax + b +2nπ

)

348/16

6. h(x) + g(x) h(n)(x) + g(n)(x)

7. k g k g(n)(x), k XHõ ã≤÷~° ã¨OYº

8. Leibnitz Theorem (hg)(x)( ) ( )

0

( ) ( )r

nn r r

cr

n h x g x−

=∑

Supportive Website



Answers


II 1. 1 22 cos 2

2n x− π +

2. 1

( 1) ! 3

(3 4)

n n

n

n

x +−

+

3. 1 1

( 2)( 1) ! 2( 1) !

( 1) ( 2)

n n

n n

n n

x x+ +− − −+

− −


I. 1. 3

16 6−ππ

2. −3 ex (sinx + cosx)

II. 1. (i) e3x+5 3n−2 [9x2 + 6nx + n(n − 1)]

(ii) eax [an x3 + 3n x2 an−1 + 3n (n−1) an−2 x + n(n−1)(n−2) an−3]

(iii) 2n−1[2x2 cos(2x + 2nπ

) + 2nx cos (2x + (n−1) π/2) + ( 1)

2n n−

cos (2x + (n−2) π/2)]

MATHEMATICS 349

Notes

MODULE - VCalculus

Integration

26

INTEGRATION

In the previous lesson, you have learnt the concept of derivative of a function. You have alsolearnt the application of derivative in various situations.Consider the reverse problem of finding the original function, when its derivative (in the form ofa function) is given. This reverse process is given the name of integration. In this lesson, we shallstudy this concept and various methods and techniques of integration.


explain integration as inverse process (anti-derivative) of differentiation;

find the integral of simple functions like nx ,sinx, cosx,

2 2 x1sec x,cosec x,secxtanx,cosec x cotx, , ex

etc.;

state the following results :

(i) f x g x dx f x dx g x dx

(ii) kf x dx k f x dx

find the integrals of algebraic, trigonometric, inverse trigonometric and exponentialfunctions;find the integrals of functions by substitution method.evaluate integrals of the type

2 2 2 2 2 2

dx dx dx, , ,

x a a x x a 22 2

dx dx, ,

ax bx ca x

22dx px q dx,

ax bx cax bx c, 2

px q dx

ax bx c

MATHEMATICS

Notes

MODULE - VCalculus

350

Integration

derive and use the result

f ' xn f x C

f x�

state and use the method of integration by parts;evaluate integrals of the type :

2 2x a dx, 2 2a x dx, axe sinbxdx , axe cosbx dx ,

2px q ax bx c dx , 1sin x dx , 1cos x dx ,

n msin x cos x dx ,dx

a b s i n x,

dxa b c o s x

derive and use the resultx xe f x f ' x dx e f x c ; and

integrate rational expressions using partial fractions.

EXPECTED BACKGROUND KNOWLEDGEDifferentiation of various functionsBasic knowledge of plane geometryFactorization of algebraic expressionKnowledge of inverse trigonometric functions

26.1 INTEGRATIONIntegration literally means summation. Consider, the problem of finding area of region ALMBas shown in Fig. 26.1.

Fig. 26.1

We will try to find this area by some practical method. But that may not help every time. Tosolve such a problem, we take the help of integration (summation) of area. For that, we dividethe figure into small rectangles (See Fig.26.2).

Fig. 26.2

MATHEMATICS 351

Notes

MODULE - VCalculus

Integration

Unless these rectangles are having their width smaller than the smallest possible, we cannot findthe area.This is the technique which Archimedes used two thousand years ago for finding areas, volumes,etc. The names of Newton (1642-1727) and Leibnitz (1646-1716) are often mentioned as thecreators of present day of Calculus.The integral calculus is the study of integration of functions. This finds extensive applications inGeometry, Mechanics, Natural sciences and other disciplines.In this lesson, we shall learn about methods of integrating polynomial, trigonometric, exponentialand logarithmic and rational functions using different techniques of integration.

26.2. INTEGRATION AS INVERSE OF DIFFERENTIATIONConsider the following examples :

(i) 2d x 2xdx

(ii)d sinx cosx

dx(iii) x xd e e

dxLet us consider the above examples in a different perspective

(i) 2x is a function obtained by differentiation of 2x .

2x is called the antiderivative of 2 x(ii) cos x is a function obtained by differentiation of sin x

sin x is called the antiderivative of cos x

(iii) Similarly, xe is called the antiderivative of xeGenerally we express the notion of antiderivative in terms of an operation. This operation iscalled the operation of integration. We write

1. Integration of 2 x is 2x 2. Integration of cos x is sin x

3. Integration of xe is xe

The operation of integration is denoted by the symbol .

Thus

1. 22 x d x x 2. cos x d x sin x 3. x xe d x e

Remember that dx is symbol which together with symbol denotes the operation of integration.

The function to be integrated is enclosed between and dx.

Definition : If d f x f ' xdx

, then f (x) is said to be an integral of 'f (x) and is written

as f '(x)dx f(x)

The function f '(x) which is integrated is called the integrand.

MATHEMATICS

Notes

MODULE - VCalculus

352

IntegrationConstant of integration

If 2 dyy x ,then 2xdx

22xdx x

Now consider 2 2d dx 2 or x cdx dx

where c is any real constant . Thus, we see that

integral of 2x is not unique. The different values of 2 x d x differ by some constant. Therefore,

22 x d x x C , where c is called the constant of integration.

Thus x xe d x e C , c o s x d x sinx c

In general f ' (x)dx f x C . The constant c can take any value.

We observe that the derivative of an integral is equal to the integrand.

Note : f(x)dx , f(y)dy , f(z)dz but not like f(z)dx

Example 26.1 Find the integral of the following :

(i) 3x (ii) 30x (iii) nxSolution :

(i)4

3 xx dx C4

, since4 3

3d x 4xx

dx 4 4

(ii)31

30 xx dx C31

, since31 30

30d x 31xx

dx 31 31

(iii)n 1

n xx dx Cn 1

sincen 1

n 1 n nd x 1 d 1x n 1 x xdx n 1 n 1 d x n 1

Example 26.2 (i) If dy cosxdx

, find y.. (ii) If dy s i n xdx

, find y..

Solution : (i)dy dx cosx dxdx

y sin x C

(ii)dy dx sinx dxdx

y cosx C

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Notes

MODULE - VCalculus

Integration

26.3 INTEGRATION OF SIMPLE FUNCTIONS

We have already seen that if f (x) is any integral of f '(x), then functions of the form f (x) + Cprovide integral of f '(x). We repeat that C can take any value including 0 and thus

f '(x)dx f ( x ) C

which is an indefinite integral and it becomes a definite integral with a defined value of C.

Example 26.3 Write any 4 different values of 34x dx .

Solution : 3 44x dx x C , where C is a constant.

The four different values of 34x dx may be 4 4 4x 1, x 2,x 3 and 4x 4 etc.

Integrals of some simple functions given below. The validity of the integrals is checked by showingthat the derivative of the integral is equal to the integrand.

Integral Verification

1.n 1

n xx dx Cn 1

∵n 1

nd xC x

dx n 1

where n is a constant and n 1.

2. sinx dx cosx C ∵d cosx C s i n x

dx

3. cosx dx sinx C ∵d sinx C cosx

dx

4. 2sec x dx tan x C ∵ 2d tan x C sec xdx

5. 2cosec x dx cotx C ∵ 2d cot x C cosec xdx

6. secx tan x dx secx C ∵d sec x C secx tan x

dx

7. cosec x cot x dx cosecx C ∵d cosecx C cosec x cot x

dx

8. 12

1dx sin x C

1 x∵

12

d 1sin x C

dx 1 x

9. 12

1dx tan x C

1 x ∵1

2d 1

tan x Cdx 1 x

10. 12

1dx sec x C

x x 1∵

12

d 1sec x C

dx x x 1

MATHEMATICS

Notes

MODULE - VCalculus

354

Integration

11. x xe dx e C ∵ x xd e C edx

12.x

x aa dx Cloga ∵

xxd a 1

C adx log a x if x > 0

13.1 dx log x Cx ∵

d log x Cdx

WORKING RULE

1. To find the integral of nx , increase the index of x by 1, divide the result by new indexand add constant C to it.

2.1 dx

f x will be very often written as dxf x

.


1. Write any five different values of 52x dx

2. Write indefinite integral of the following :

(a) 5x (b) cos x (c) 0

3. Evaluate :

(a) 6x dx (b) 7x dx (c)1 dxx

(d) x x3 5 dx

(e) 3 x dx (f) 9x dx (g)1

dxx (h) 9 8x dx

4. Evaluate :

(a) 2cos

dsin

(b) 2sin

dcos

(c)2 2

2cos sin

dcos

(d) 21

dsin

26.4 PROPERTIES OF INTEGRALSIf a function can be expressed as a sum of two or more functions then we can write the integral

of such a function as the sum of the integral of the component functions, e.g. if 7 3f (x) x x ,then

MATHEMATICS 355

Notes

MODULE - VCalculus

Integration

7 3f x dx x x dx

7 3x dx x dx

8 4x x C8 4

So, in general the integral of the sum of two functions is equal to the sum of their integrals.

f x g x dx f x dx g x dx

Similarly, if the given function

7 2f x x x

we can write it as 7 2f x dx x x dx

7 2x dx x dx

8 3x x C8 3

The integral of the difference of two functions is equal to the difference of their integrals.

i.e. f x g x dx f x dx g x dx

If we have a function f(x) as a product of a constant (k) and another function g x

i.e. f(x) kg(x) , then we can integrate f(x) as

f x dx kg x dx

k g x dx

Integral of product of a constant and a function is product of that constant and integral of thefunction.

i.e. kf x dx k f x dx


(ii) x4 dx (ii) x x2 3 dx

Solution :(i)x

x 44 dx Clog4

(ii)x

x xx

22 3 dx dx

3

MATHEMATICS

Notes

MODULE - VCalculus

356

Integration

x2 dx3

x23 C

2log3

Remember in (ii) it would not be correct to say that

x x x x2 3 dx 2 dx 3 dx

Because

x

x xx x

22 3 32 dx 3 dx C C

2log 2 log3 log3

Therefore, integral of a product of two functions is not always equal to the product of theintegrals. We shall deal with the integral of a product in a subsequent lesson.


(i) ndx

cos x, when n = 0 and n = 2 (ii)

2 2

2sin cos

dsin

Solution :

(i) When n = 0, n odx dx

cos x cos xdx dx1

Now dx can be written as ox dx .

odx x dx

o 1x C x C0 1

When n = 2,

n 2dx dx

cos x cos x2sec x dx

tanx C

(ii)2 2

2sin cos

dsin 2

1d

sin

2cosec d

cot C

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Notes

MODULE - VCalculus

Integration


(i) s inx cosx dx (ii)2

3x 1

dxx

(iii)1 x

dxx (iv) 2 2

1 1dx

1 x 1 x

Solution : (i) s inx cosx dx sinxdx cosx dx cosx sinx C

(ii)2 2

3 3 3x 1 x 1

dx dxx x x

31 1

dx dxx x

3 1xlog x C3 1

21

log x C2x

(iii)1 x 1 x

dx dxx x x

1 12 2x x dx

3 / 222 x x C3

(iv) 2 22 2

1 1 dx dxdx

1 x 1 x1 x 1 x1 1tan x sin x C


(i) 1 sin2 d (ii)x

2

34e dx

x x 1

(iii) 2tanx cot x dx (iv)6

2x 1

dxx 1

Solution : (i) 2 21 sin 2 cos sin 2sin cos

2 2sin cos 1∵

MATHEMATICS

Notes

MODULE - VCalculus

358

Integration

2cos sin

cos sin(sign is selected depending upon the value of )(a) If 1 sin2 cos sin

then 1 sin2 d cos sin d

cos d sin d

sin cos C

(b) If 1 sin2 d cos sin d

cos d sin d

sin cos C

(ii)x x

2 2

3 34e dx 4e dx dx

x x 1 x x 1

x2

dx4 e dx 3

x x 1x 14e 3sec x C

(iii) 2 2 2tanx cot x dx tan x cot x 2tanx cot x dx

2 2tan x cot x 2 dx

2 2tan x 1 cot x 1 dx

2 2sec x cosec x dx

2 2sec xdx cosec x dx

tanx cot x C

(iv)6

4 22 2

x 1 2dx x x 1 dx

x 1 x 1 (dividing 6x 1 by 2x 1)

4 22dx

x dx x dx dx 2x 1

5 31x x x 2 tan x C

5 3Example 26.8 Evaluate :

(i)31x dx

x(ii)

5x 4x

3x4e 9e 3

dxe

(iii)dx

x a x b

MATHEMATICS 359

MODULE - VCalculus

Notes

Integration

Solution:

(i)3

1x dx

x

+ ∫ =3/ 2

3/ 2

1 1 1x 3x 3 x dx

x xx

+ + + ∫

= 3/ 23/ 2

1 dxx dx 3 x dx 3 dx

xx+ + +∫ ∫ ∫ ∫

=5/ 2 3/ 2 1/ 2

5 3 12 2 2

x x x 23 3 C

x+ + − +

=5 3 1 12 2 2 2

2x 2x 6x 2x C

5+ + − +

(ii)5x 4x 5x 4x

3x 3x 3x 3x

4e 9e 3 4e 9e 3dxdx dx dx

e e e e

− − = − −

∫ ∫ ∫ ∫

= 2x x 3x4 e dx 9 e dx 3 e dx−− −∫ ∫ ∫ = 2e2x − 9ex + e−3x + C

(iii)dx

x a x b+ + +∫Let us first rationalise the denominator of the integrand.

1 1 x a x b

x a x b x a x b x a x b

+ − += ×+ + + + + + + − +

x a x b

(x a) (x b)

+ − +=+ − +

x a x b

a b

+ − +=−

∴dx x a x b

a bx a x b

+ − +=−+ + +∫

x a x bdx dx

a b a b

+ += −− −∫ ∫

3 32 21 (x a) 1 (x b)

C3 3a b a b2 2

+ += − +− −

3 32 2

2(x a) (x b) C

3(a b) = + − + + −

360 MATHEMATICS

MODULE - VCalculus

Notes

Integration


1. Evaluate :

(a)1

x dx2

+ ∫ (b) 2

2

xdx

1 x

−+∫

(c)9 1

10x x dxx

− + ∫ (d)2 4 6

6

5 3x 6x 7x 8xdx

x

+ − − −

∫

(e)4

2

xdx

1 x+∫ (f)

22

x dxx

+ ∫

2. Evaluate :

(a)dx

1 cos 2x+∫ (b) 2tan xdx∫ (c) 2

2cos xdx

sin x∫

(d)dx

1 cos 2x−∫ (e) 2

sin xdx

cos x∫ (f) (cosec x cot x) cos ec x dx−∫3. Evaluate :

(a) 1 cos 2x dx+∫ (b) 1 cos 2x dx−∫ (c)1

dx1 cos 2x−∫

4. Evaluate :

(a) x 2 dx+∫ (b) 17

17dx

(x 1)+

We know that

n 1 nd(ax b) (n 1)a(ax b)

dx++ = + +

⇒n n 11 d

(ax b) (ax b) , n 1(n 1)a dx

++ = + ≠ −+

n 1n (ax b)

(ax b) dx C(n 1)a

+++ = ++∫

To find the integral of (ax + b)n, increase the index by one and divide the result by the increasedindex and the coefficient of x and add a constant of integration.

Example 26.9 Integrate the following :

(i) (x + 9)11 (ii) ex+7 (iii) cos (x + 5π)

(vi) cosec2(2x + 3)

MATHEMATICS 361

MODULE - VCalculus

Notes

Integration

Solution :

(i)11 1

11 (x 9)(x 9) dx C

11 1

+++ = ++∫

12(x 9)C

12

+= +

(ii) x 7 x 7e dx e . e dx+ =∫ ∫7 xe e dx= ∫ (e7 is a constant quantity)

= e7 . ex + C = ex+7 + C

(iii) cos(x 5 )dx sin(x 5 ) C+ π = + π +∫d d

(sin(x 5 ) C) cos(x 5 ). (x 5 ) cos(x 5 )dx dx

+ π + = + π + π = + π �

(iv) 2 cot(2x 3)cosec (2x 3)dx C

2

− ++ = +∫

22d cot(2x 3) cosec (2x 3)

C 2 cosec (2x 3)dx 2 2

− + − + + = − = + �


3(2x 5)(x 1)x dx−

+ −∫Solution :

23(2x 5)(x 1)x dx−

+ −∫2

32(2x 3x 5)x dx−

= + −∫2 2 23 3 3

2 12x 3x 5x dx

− −− − = + − ∫4 / 3 1/ 3 2 / 32 x dx 3 x dx 5 x dx−= + −∫ ∫ ∫

4 1 21 1 1

3 3 3x x x2 3 5 C

4 1 21 1 1

3 3 3

+ + − +

= + − ++ + − +

7 4 1

3 3 36 9x x 15x C

7 4= + − +

Example 26.11 Evaluate : 2 x 2

2

x e (x 1)dx

x

+ +∫

362 MATHEMATICS

MODULE - VCalculus

Notes

Integration

Solution :2 x 2

2

x e (x 1)dx

x

+ +∫ =

2 x 2

2

x e x 2x 1dx

x

+ + +∫

= x

2

2 1e 1 dx

x x + + + ∫

=x

2

dx dxe dx dx 2

x x+ + +∫ ∫ ∫ ∫

=x 1

e x 2log | x | Cx

+ + − +

Example 26.12 Integrate sec2x cosec2x w.r.t. x

Solution : 2 2sec x cosec x dx∫ = 2 2(1 tan x)(1 cot x) dx+ +∫2 2 2 2sec x 1 tan x and cosec x 1 cot x = + = + �

= 2 2 2 2(1 tan x cot x tan x cot x) dx+ + +∫= 2 2(1 tan x) (1 cot x)dx+ + +∫= 2 2(sec x cosec x)dx+∫= tan x − cot x + C

Example 26.13 Evaluate : 2 2

7 6dx

1 x 1 x

− − + − ∫

Solution : 2 22 2

7 6 dx dxdx 7 6

1 x 1 x1 x 1 x

− − = − − + +− − ∫ ∫ ∫

Example 26.14 Evaluate : (x cos x)dx+∫Solution : (x cos x) dx x dx cos x dx+ = +∫ ∫ ∫

2xsin x C

2= + +

26.5 TECHNIQUES OF INTEGRATION

26.5.1 Integration By Substitution

This method consists of expressing f (x) dx∫ in terms of another variable so that the resultant

MATHEMATICS 363

Notes

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Integration

function can be integrated using one of the standard results discussed in the previous lesson.First, we will consider the functions of the type f ax b , a 0 where f (x) is a standardfunction.


(i) sin ax b dx (ii) cos 7x dx4 (iii)

xsin dx

4 2

Solution : (i) sin ax b dx

Put ax b t .

Thendtadx

ordtdxa

dtsin ax b dx sin ta

(Here the integration factor will be replaced by dt.)

1 sint dta1 cost Ca

cos ax bC

a

(ii) cos 7x dx4

Put 7x t4 7dx dt

dtcos 7x dx cos t

4 7

1 cost dt71 sin t C7

1sin 7x C

7 4

(iii)x

sin dx4 2

Putx t

4 2

MATHEMATICS

Notes

MODULE - VCalculus

364

Integration

Then1 dt2 dx

or dx 2dt

xsin dx 2 sint dt

4 2

2 cost C

2cos t C

x2cos C

4 2

Similarly, the integrals of the following functions will be—

sin2x dx =1 cos2x C2

sin 3x dx3 =

1cos 3x C

3 3

xsin dx

4 4 =x

4cos C4 4

cos ax b dx =1 sin ax b Ca

cos2x dx =1 sin 2x C2


(i) nax b dx , where n 1 (ii)1 dx

ax b

Solution : (i) nax b dx , where n 1

Put ax b tdtadx

or dtdxa

n n1ax b dx t dta

n 11 t Ca n 1

n 1ax b1 Ca n 1

where n 1

MATHEMATICS 365

Notes

MODULE - VCalculus

Integration

(ii)1 dx

ax b

Put ax b t1dx dta

1 1 dtdxax b a t

1 log t Ca1 log ax b Ca


(i) 5x 7e dx (ii) 3x 3e dx

Solution : (i) 5x 7e dx

Put 5x 7 tdtdx5

5x 7 t1e dx e dt5

t1 e C5

5x 71 e C5

(ii) 3x 3e dx

Put 3x 3 t1dx dt3

3x 3 t1e dx e dt3

t1 e C3

3x 31 e C3

MATHEMATICS

Notes

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366

Integration

Likewise ax b ax b1e dx e Ca

Similarly, using the substitution ax + b = t, the integrals of the following functions will be :

nax b dxn 1ax b1 C, n 1

a n 1

1 dxax b

1 log ax b Ca

sin ax b dx1 cos ax b Ca

cos ax b dx1 sin ax b Ca

2sec ax b dx1 tan ax b Ca

2cosec ax b dx1 cot ax b Ca

sec ax b tan ax b dx1 sec ax b Ca

cosec ax b cot ax b dx1 cosec ax b Ca


(i) 2sin x dx (ii) 3sin x dx (iii) 3cos x dx (iv) sin3x sin2x dx

Solution : We use trigonometrical identities and express the functions in terms of sines andcosines of multiples of x

(i) 2 1 cos2xsin x dx dx2

2 1 cos2xsin x

2∵

1 1 cos2x dx21 11dx cos2x dx2 21 1x sin2x C2 4

(ii) 3 3sinx sin3xsin x dx dx4

3sin3x 3sinx 4sin x∵

1 3sinx sin 3 x dx4

MATHEMATICS 367

Notes

MODULE - VCalculus

Integration

1 cos3x3cosx C

4 3

(iii) 3 cos3x 3cos xcos x dx dx4

3cos3x 4cos x 3cosx∵

1 cos3x 3cosx dx4

1 sin3x3sinx C

4 3

(iv)1sin3x sin2x dx 2sin3x sin2x dx2

2sinAsinB cos A B cos A B∵

1 cosx cos5x dx2

1 sin5xsinx C

2 5

CHECK YOUR PROGRESS 26.31. Evaluate :

(a) sin 4 5x dx (b) 2sec 2 3x dx

(c) sec x dx4 (d) cos 4x 5 dx

(e) sec 3x 5 tan 3x 5 dx

(f) cosec 2 5x cot 2 5x dx

2. Evaluate :

(a) 4dx

3 4x (b) 4x 1 dx (c) 104 7x dx

(d) 34x 5 dx (e)1 dx

3x 5(f)

1dx

5 9x

(g) 22x 1 dx (h)1 dx

x 13. Evaluate :

(a) 2x 1e dx (b) 3 8xe dx (c) 7 4x1 dx

e

MATHEMATICS

Notes

MODULE - VCalculus

368

Integration4. Evaluate :

(a) 2cos xdx (b) 3 3sin xcos x dx

(c) sin4x cos3x dx (d) cos4x cos2x dx

26.5.2 Integration of Function of The Type f ' xf x

To evaluate f ' x

dxf x

, we put f (x) = t. Then f ' (x) dx = dt.

f ' x dtdxf x t

log t C

log f x C

Integral of a function, whose numerator is derivative of the denominator, is equal to the logarithmof the denominator.


(i) 22x

dxx 1

(ii)3

44x

dx5 x

(iii)dx

2 x 3 x

Solution :

(i) Now 2x is the derivative of 2x 1.By applying the above result, we have

222x

dx log x 1 Cx 1

(ii) 34x is the derivative of 45 x .

34

44x

dx log 5 x C5 x

(iii)1

2 x is the derivative of 3 x

dxlog 3 x C

2 x 3 x


(i)x x

x xe e

dxe e

(ii)2x

2xe 1

dxe 1

MATHEMATICS 369

Notes

MODULE - VCalculus

IntegrationSolution :

(i) x xe e is the derivative of x xe ex x

x xx x

e edx log e e C

e eAlternatively,

Forx x

x xe e

dxe e

,

Put x xe e t.

Then x xe e dx dt

x x

x xe e dt

dxte e

log t C

x xlog e e C

(ii)2x

2xe 1

dxe 1

Here 2xe 1 is not the derivative of 2xe 1. But if we multiply the numerator and denominator

by xe , the given function will reduce tox x

x xe e

dxe e

x xlog e e C

2x x xx x

2x x xe 1 e e

dx log e e Ce 1 e e

x x x xe e is the derivative of e e∵


(a) 2x

dx3x 2

(b) 22x 1

dxx x 1

(c) 22x 9

dxx 9x 30

(d)2

3x 1

dxx 3x 3

(e) 22x 1

dxx x 5

(f)dx

x 5 x

(g)dx

x 8 log x

MATHEMATICS

Notes

MODULE - VCalculus

370

Integration2. Evaluate :

(a)x

xe

dx2 be

(b) x xdx

e e

26.5.3 Integration by Substitution

Example 26.21

(i) tanxdx (ii) secxdx (iii)1 t a n x dx1 t a n x

(iv)1 s i n x

dx1 cosx (v) 5cosec x cotxdx (vi)

sin x dxsin x a

Solution :

(i)s i n xtanxdx dxcosx

s i n x dxcosx

log cos x C ( sin x∵ is derivative of cos x)

1log C

cosx or log sec x C

tanxdx log sec x C

Alternatively,

sin x dxtanxdxcosx

sinx dxcosx

Put cosx t.Then sinx dx dt

dttanxdxt

log t C

log cosx C

1log C

cosx

log secx C

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Notes

MODULE - VCalculus

Integration

(ii) sec x dx

sec x can not be integrated as such because sec x by itself is not derivative of any function. But

this is not the case with 2sec x and sec x tan x. Now sec x dx can be written as

sec x tan xsecx dx

sec x tan x

2sec x sec x tan xdx

secx t a n xPut sec x tan x t.

Then 2sec x tan x sec x dx dt

dtsecx dxt

log t C

log secx tan x C

(iii)1 tanx cosx s i n xdx dx1 tan x cosx s i n x

Put cosx sinx tSo that sinx cosx dx dt

1 tanx 1dx dt1 tan x t

log t C

log cosx sinx C

(iv)1 sinx 1 s i n xdt dx dx1 cosx 1 cosx 1 cosx

2 2

x x2sin cos1 2 2dxx x2cos 2cos2 2

21 x xsec dx tan dx

2 2 2

Putx t2

1 dx dt2

21 s i n x dx sec td t 2 tant dt1 cosx

tant 2log cost C

MATHEMATICS

Notes

MODULE - VCalculus

372

Integration

=x x

tan 2log cos C2 2

(v) 5 4cosec x cotx dx cosec xcosec x cot x dx

Put cosec x t .Then cosec x cot x dx dt

5 4cosec x cot x dx t dt

5t C5

5cosec xC

5

(vi)sin x dx

sin x aPut x a tThen dx dt and x t a

sin t asin x dx dtsin x a s in t

s intcosa cos t s ina dts i n t

sin A B sinAcosB cosAsinB∵

cosa dt sina cott dt

(cos a and sin a are constants.)cosa t sinalog sint C

x a cosa sina log sin x a C

Example 26.22 Evaluate 2 21

dxa x

Solution : Put x a sin dx acos d

2 2 2 2 21 acos

dx da x a a sin

21 cos

da 1 sin

1 1 da cos

MATHEMATICS 373

Notes

MODULE - VCalculus

Integration

1 sec da

1 log sec tan Ca

1 1 sinlog C

a cos

2

2

x11 alog Ca x1

a

2 2

1 a xlog C

a a x

1 a xlog C

a a x

121 a xlog C

a a x

1 a xlog C

2a a x

Example 26.23 Evaluate : 2 21

dxx a

Solution : Put x a sec dx asec tan d

2 2 2 2 21 asec tan d

dxx a a sec a

21 sec tan

da tan

2 2tan sec 1

1 sec da tan

1 1 1d cosec da sin a

1 log cosec cot Ca

MATHEMATICS

Notes

MODULE - VCalculus

374

Integration

1 1 coslog C

a sin

2

2

a11 xlog Ca a1

x

2 2

1 x alog C

a x a

1 x alog C

a x a

1 x alog C

2a x a

Example 26.24 2 21

dxa x

Solution : Put x a tan 2dx a sec d

2

2 2 2 21 a secdx d

a x a 1 tan

1 da

1 Ca

xtan

a1 x

tana

11 xtan Ca a

Example 26.25 2 2

1dx

a x

Put x a sin dx a cos d

2 2 2 2 2

1 acosdx d

a x a a sin

acos dacos

d

MATHEMATICS 375

Integration

MODULE - VCalculus

Notes

= θ + C

=1 x

sin Ca

− +

Example 26.262 2

1dx

x a−∫

Solution : Let x = a sec θ ⇒ dx = a sec θ tan θ dθ

∴ 2 2 2

1 a sec tan

x a a sec 1

θ θ=− θ −

∫ ∫

sec d= θ θ∫ = log |sec θ + tan θ| + C

2 2x 1log x a C

a a= + − +

2 2log x x a C= + − +

Example 26.272 2

1dx

a x+∫

Solution : Put x = a tan θ dx = a sec2 θ dθ

= sec dθ θ∫ [As example 26.25]

= log |sec θ + tan θ| + C

=2 21 x

log a x Ca a

+ + +

= 2 2log a x x C+ + +

Example 26.28 2

4

x 1dx

x 1

++∫

Solution : Since x2 is not the derivative of x4 + 1, therfore, we write the given integral as

2

22

11

x dx1

xx

+

+∫

Let1

x tx

− =

∴ 2

11 dx dt

x + =

376 MATHEMATICS

Integration

MODULE - VCalculus

Notes

Also2 2 2 2

2 2

1 1x 2 t x t 2

x x− + = ⇒ + = +

∴2

22

2

11

dtx dx1 t 2xx

+=

++∫ ∫

2 2

dt

(t) ( 2)=

+∫

11 ttan C

2 2−= +

1

1x1 xtan C

2 2−

− = +

Example 26.292

4

x 1dx

x 1

−+∫

Solution :

2 2

42

2

11

x 1 xdx dx1x 1 xx

−− =+ +

∫ ∫

Put1

x t.x

+ =

Then 2

11 dx dt

x − =

Also2 2

2

1x 2 t

x+ + =

⇒2 2

2

1x t 2

x+ = −

∴2

22

2

11

dtx dx1 t 2xx

−=

−+∫ ∫

2 2

dt

(t) ( 2)=

−∫

1 t 2log C

2 2 t 2

−= ++

MATHEMATICS 377

Notes

MODULE - VCalculus

Integration

1x 21 xlog C12 2 x 2x

Example 26.302

4x

dxx 1

Solution : In order to solve it, we will reduce the given integral to the integrals given in Examples26.28 and 26.29.

i.e.,2 2 2

4 4 4x 1 x 1 x 1

dx dx2x 1 x 1 x 1

2 2

4 41 x 1 1 x 1

dx dx2 2x 1 x 1

1

1 1x x 21 1 1x xtan log C12 2 2 2 x 2x

Example 26.31 41

dxx 1

Solution : We can reduce the given integral to the following form

2 2

4

x 1 x 11 dx2 x 1

2 2

4 41 x 1 1 x 1

dx dx2 2x 1 x 1

1

1 1x x 21 1 1x xtan log C12 2 2 2 2 x 2x

Example 26.32 (a) 21

dxx x 1

(b)2

4 2x 1

dxx x 1

Solution : (a) 2 2

1 1dx dx1 1x x 1 x x 14 4

21 dx

1 3x2 4

MATHEMATICS

Notes

MODULE - VCalculus

378

Integration

22

1 dx1 3x2 2

1

1x1 2tan C3 3

2 2

(b)2 2

4 2 22

11x 1 xdx dx1x x 1 x 1x

Put1x t.x 2

11 dx dt

x

Also 2 22

1x 2 t

x

2 22

1x 1 t 1

x

222

2

11 dtx dx1 t 1x 1x

1 t 1log C

2 t 1

1x 11 xlog C12 x 1x

Example 26.33 tan x dx

Solution : Let 2tan x t 2sec xdx 2t dt

22t

dx dtsec x

42t

dt1 t

MATHEMATICS 379

Notes

MODULE - VCalculus

Integration

42t

tanxdx t dt1 t

2

42t

dt1 t

2 2

4 4t 1 t 1

dtt 1 t 1

2 2

4 4t 1 t 1

dt dtt 1 t 1

Proceed according to example 26.28 and Example 26.29 solved before.

Example 26.34 cot x dx

Solution : Let 2cot x t 2cosec x dx 2t dt

22t

dx dtcosec x

42t

dtt 1

42t

cot x dx t dtt 1

2

42t

dtt 1

2 2

4 4t 1 t 1

dtt 1 t 1

Proceed according to Examples 26.28 and 26.29 solved before.

Example 26.35 tan x cot x dx

Let sinx cosx t cosx sin x dx dt

Also 21 2sinxcosx t21 t 2s inx cosx

21 t sinx cosx2

MATHEMATICS

Notes

MODULE - VCalculus

380

Integration

2

sin x cosx dtdxcosxs inx 1 t

2

2

dt2

1 t

12 sin sinx cosx C

(Using the result of Example 26.25)


(a) 2

dx

8 3x x(b)

dxx 1 2x

Solution :

(a) 2 2

dx dx

8 3x x 8 x 3x

2

dx9 98 x 3x4 4

2 2

dx

41 3x2 2

1

3x2sin C

412

1 2x 3sin C

41

(b)2

dx dxx 1 2x x 2x

2

1 dx2 x x

2

2

1 dx2 1 x 1x

16 2 16

MATHEMATICS 381

Notes

MODULE - VCalculus

Integration

2 21 dx2 1 1x

4 4

1

1x1 4sin C124

11sin 4x 1 C

2


(a)2

2x

dxx 9

(b)x

2xe

dxe 1

(c) 4x

dx1 x

(d) 2

dx

16 9x(e) 2

dx1 3sin x

(f) 2

dx

3 2x x

(g) 2dx

3x 6x 21(h) 2

dx

5 4x x(i) 2

dx

x 3x 12

(j) 4 4d

sin cos(k)

x

2x

e dx

1 e(l)

1 x dx1 x

(m) 2

dx

2ax x(n)

2

6

3xdx

9 16x(o) 2

x 1dx

x 1

(p) 2

dx

9 4x(q) 2

sind

4cos 1(r)

2

2

sec xdx

tan x 4

(s) 21 dx

x 2 1 (t) 2

1dx

16x 25

26.6 INTEGRATION BY PARTS

In differentiation you have learnt thatd d dfg f g g f

dx dx dx

382 MATHEMATICS

MODULE - VCalculus

Notes

Integration

ord d d

f (g) (fg) g (f )dx dx dx

= − (1)

Also you know thatd

(fg)dx fgdx

=∫Integrating (1), we have

df (g)dx

dx∫ = d d

(fg)dx g (f )dxdx dx

−∫ ∫d

fg g (f )dxdx

= −∫ (2)

if we take f = d

u(x); (g) v(x),dx

=

(2) becomes u(x)v(x)dx∫d

u(x) . v(x)dx (u(x)) v(x)dx dxdx

= − ∫ ∫ ∫

= I function × integral of II function − [differential coefficient of Ifunction integral of II function]dx×∫A B

Here the important factor is the choice of I and II function in the product of two functionsbecause either can be I or II function. For that the indicator will be part 'B' of the result above.

The first function is to be chosen such that it reduces to a next lower term or to a constant termafter subsequent differentiations.

Inequations of integration like

x sin x, x cos2 x, x2 ex

(i) algebraic function should be taken as the first function.

(ii) If there is no algebraic function then look for a function which simplifies the production in'B' as above; the choice can be in order of preference like choosing first function.

(i) an inverse function (ii) a logarithmic function

(iii) a trigonometric function (iv) an exponential function

The following example will give a practice to the concept of choosing first function.

I function II function

1. x cos x dx∫ x (being algebraic) cos x

2. 2 xx e dx∫ x2 (being algebraic) ex

3. 2x log x dx∫ log x x2

MATHEMATICS 383

MODULE - VCalculus

Notes

Integration

I function II function

4. 2

log x

(1 x )+∫ logx 2

1

(1 x)+

5. 1x sin x dx−∫ sin−1x x

6. log x dx∫ log x 1

(In single function of logarithm and inverse trigonometric wetake unity as II function)

7. sin−1 x dx sin−1x 1


x cos x dx∫ Solution : Taking the polynomial (algebraic function) x as the first function and trigonometricfunction cos x as the second function, we get

dx cos x dx x cos x dx (x) . cos x dx dx

dx = − ∫ ∫ ∫ ∫

I II

dI function Integral of II function (I function). (II fuction dx) dx

dx

× − ∫ ∫

= x sin x 1 . sin x dx− ∫= x sin x − [− cos x] + C

= x sin x + cos x + C Example 26.38 Evaluate :

2x sin x dx∫Solution : Taking algebraic function x2 as I function and sin x as II function, we have,

2 2 2dx sin x dx x sin x (x ) sin x dx dx

dx = − ∫ ∫ ∫ ∫

I II2x cos x 2 x( cos x)dx= − − −∫2x cos x 2 x cos x dx= − + ∫ (1)

Again, x cos x dx x sin x cos x C= + +∫ (2)

Substituting (2) in (1) we have

384 MATHEMATICS

MODULE - VCalculus

Notes

Integration

2 2x sin x dx x cos x 2[x sin x cos x] C= − + + +∫ = −x2 cosx + 2x sinx + 2 cos x + C

Example 26.39 Evaluate :2x log x dx∫

Solution : In order of preference log x is to be taken as I function.

∴ 2log x x dx∫ =3 3x 1 x

log x . dx3 x 3

− ∫ I II

=3 2x x

log x dx3 3

− ∫

=3 2x 1 x

log x C3 3 3

− +

=3 3x x

log x C3 9

− +


2

log xdx

(1 x)+∫

Solution : 2

log xdx

(1 x)+∫ 2

1log x . dx

(1 x)=

+∫I II

1 1 1log x dx

1 x x 1 x

− = − − × + + ∫

log x 1dx

1 x x(1 x)

−= ++ +∫

log x 1 1dx

1 x x 1 x

− = + − + + ∫

log x 1 1dx dx

1 x x 1 x

−= + −+ +∫ ∫

log xlog | x | log |1 x | C

1 x

−= + − + ++

log x xlog C

1 x 1 x

−= + ++ +

MATHEMATICS 385

MODULE - VCalculus

Notes

Integration


2xx e dx∫

Solution : 2xx e dx∫2x 2xe e

x 1. dx2 2

= −∫2x 2xe 1 e

x C2 2 2

= − +

2x

2xe 1x e C.

2 4= − +

Example 26.42 Evaluate :1sin x dx−∫

Solution : 1sin x dx−∫ = 1sin x . 1 .dx−∫

=1

2

xx sin x dx

1 x

− −−

∫Let 1 − x2 = t

⇒ −2x − dx = dt

⇒ 1

x dx dt2

−=

∴ 2

x 1 dtdx

2 t1 x= −

−∫ ∫

t C= − +

21 x C= − − +

∴ 1 1 2sin x dx x sin x 1 x C− −= + − +∫


Evaluate :

1. (a) x sin x dx∫ (b) 2(1 x )cos 2x dx+∫ (c) x sin 2x dx∫2. (a) 2x tan x dx∫ (b) 2 2x sin x dx∫3. (a) 3x log 2x dx∫ (b) (1 − x2)logx dx (c) 2(log x) dx∫

4. (a) n

log xdx

x∫ (b)log(log x)

dxx∫

386 MATHEMATICS

MODULE - VCalculus

Notes

Integration

5. (a) 2 3xx e dx∫ (b) 3xx e dx∫6. 2x (log x) dx∫7. (a) 1sec x dx−∫ (b) 1x cot x dx−∫

26.7 INTEGRAL OF THE FORM xe [f (x) f (x)]dx′+∫xe [f (x) f (x)]dx′+∫

Where f '(x) is the differentiation of f(x). In such type of integration while integrating by partsthe solution will be ex (f (x)) + C.

For example, consider

xe [tan x log sec x]dx+∫ (1)

Let f(x) = log sec x,

thensec x tan x

f '(x) tan xsec x

= =

So (1) can be rewritten as

x x xe [f '(x) f (x)]dx e (f (x)) C e log sec x C+ = + = +∫Alternatively, you can evaluate it as under :

x x xe [tan x log sec x]dx e tan x dx e log sec x dx+ = +∫ ∫ ∫I II

x x xe log sec x e log sec x dx e log sec x dx= − +∫ ∫= ex log sec x + C

Example 26.43 Evaluate the following :

(a)x

2

1 1e dx

x x − ∫ (b)

x 1 x log xe dx

x

+ ∫

(c)x

2

x edx

(x 1)+∫ (d)x 1 sin x

e dx1 cos x

+ + ∫

Solution :

(a) x x x2

1 1 1 d 1 1e dx e dx e

x x dx x xx

− = + = ∫ ∫

(b) x x1 x log x 1e dx e log x dx

x x

+ = + ∫ ∫

MATHEMATICS 387

MODULE - VCalculus

Notes

Integration

x de log x (log x) dx

dx = + ∫

= ex log x + C

(c)x

x2 2

x e x 1 1dx e dx

(x 1) (x 1)

+ −=+ +∫ ∫

x2

1 1e dx

x 1 (x 1)

= − + +

∫

x 1 d 1e dx

x 1 dx (x 1)

= + + +

∫

x 1e C

x 1 = + +

(d) x x

2

x x1 2sin cos1 sin x 2 2e dx e dx

x1 cos 2cos2

+ + = +

∫ ∫

x 21 x xe sec tan dx

2 2 2 = + ∫

x x d xe tan tan dx

2 x 2

= + ∫

x xe tan C

2= +


(a) 3sec x dx∫ (b) xe sin x dx∫Solution :

(a) 3sec x dx∫Let 2I sec x sec x dx= −∫

sec x . tan x sec x tan x . tan x dx= −∫∴ 3I sec x tan x (sec x sec x ) dx= − −∫ ( )2 2tan x sec x 1= −�

or 3I sec x tan x sec x dx sec x dx= − +∫ ∫or 2I sec x tan x sec x dx= +∫

388 MATHEMATICS

MODULE - VCalculus

Notes

Integration

or I = sec x tan x + log |sec x + tan x| + C1

or I = 1

2[sec x tan x + log |sec x + tan x|] + C

(b) xe sin x dx∫Let xI e sin x dx= ∫

= x xe ( cos x) e ( cos x) dx− − −∫= x xe cos x e cos x dx− +∫= ( )x x xe cos x e sin x e sin x dx− + −∫

∴ I = −ex cos x + ex sin x − Ior 2I = −ex cos x + ex sin x

or I = xe

(sin x cos x) C2

− +


2 2a x dx−∫Solution :

Let 2 2 2 2I a x dx a x . 1 dx= − = −∫ ∫Integrating by parts only and taking 1 as the second function, we have

( )2 2

2 2

1I a x x ( 2x) . x dx

2 a x= − − −

−∫

22 2

2 2

xx a x dx

2 a x= − +

−∫

2 2 22 2

2 2

a (a x )x a x dx

a x

− −= − +−

∫

2 2 2 2 2

2 2

1x a x a dx a x dx

a x= − + − −

−∫ ∫

∴ I = 2 2 2 1 x

x a x a sin Ia

− − + −

or 2I = 2 2 2 1 x

x a x a sina

− − +

MATHEMATICS 389

MODULE - VCalculus

Notes

Integration

or I = 2 2 2 11 x

x a x a sin C2 a

− − + +

Similarly,2 2 2

2 2 2 2x x a ax a dx log x x a C

2 2

−− = − + − +∫

∴2 2 2

2 2 2 2x a x aa x dx log x a x C

2 2

+− = − + − +∫ Example 26.46 Evaluate :

(a) 216x 25 dx+∫ (b) 216 x dx−∫ (c) 21 x 2x dx+ −∫Solution :

(a)2

2 2 225 516x 25 dx 4 x dx 4 x dx

16 4 + = + = + ∫ ∫ ∫

Using the formula for 2 2(x a ) dx+∫ we get,

2 2 2x 25 25 2516x 25 dx x log x x C

2 16 32 16

+ = + + + + +

∫

2 2x 25

16x 25 log 4x 16x 25 C8 8

= + + + + +

(b) Using the formula for 2 2(a x ) dx−∫ we get,

2 2 216 x dx (4) x dx− = −∫ ∫

2 1x 16 x

16 x sin C2 2 4

−= − + +

(c) 2 21 x

1 x 2x dx 2 x dx2 2

+ − = + −∫ ∫

21 x 1 12 x dx

2 2 16 16

= − − + + ∫

2 23 1

2 x dx4 4

= − − ∫

21

1 1x x9 1 94 42 x sin C

32 16 4 16 24

−

− − = − − + + ×

390 MATHEMATICS

MODULE - VCalculus

Notes

Integration

2 14x 1 1 9 4x 12 . 1 x 2x sin C

8 32 32−− − = + − + +

2 14x 1 9 2 4x 11 x 2x sin C

8 32 3−− −= + − + +

CHECK YOUR PROGRESS 26.6Evaluate :

1. (a) xe sec x[1 tan x ]dx+∫ (b) xe [sec x log | sec x tan x |]dx+ +∫

2. (a)x

2

x 1e dx

x

−∫ (b)

x 1

2

1e sin x dx

1 x

− − −

∫

3.x

3

(x 1)e dx

(x 1)

−+∫ 4.

x

2

xedx

(x 1)+∫

5.x sin x

dx1 cos x

++∫ 6. xe sin 2x dx∫

26.8 INTEGRATION BY USING PARTIAL FRACTIONS

By now we are equipped with the various techniques of integration.

But there still may be a case like2

4x 5,

x x 6

++ −

where the substitution or the integration by parts

may not be of much help. In this case, we take the help of another technique called techniqueof integration using partial function.

Any proper rational fraction p(x)

q(x) can be expressed as the sum of rational functions, each

having a single factor of q(x). Each such fraction is known as partial fraction and the processof obtaining them is called decomposition or resolving of the given fraction into partial fractions.

For example, 2

3 5 8x 7 8x 7

x 2 x 1 (x 2)(x 1) x x 2

+ ++ = =+ − + − + −

Here3 5

,x 2 x 1+ − are called partial fraction of

2

8x 7

x x 2

++ −

.

Iff (x)

g(x) is a proper fraction and g(x) can be resolved into real factors then,

(a) correspnding to each non repeated linear factor ax + b, there is a partial fraction of theform

MATHEMATICS 391

MODULE - VCalculus

Notes

Integration

(b) for (ax + b)2 we take the sum of two partial fractions as

2

A B

ax b (ax b)+

+ +

For (ax + b)3 we take the sum of three partial fraction as

2 3

A B C

ax b (ax b) (ax b)+ +

+ + +

and so on.

(c) For a non - fractorisable quadratic polynomial ax2 + bx + c there is a partial fraction

2

Ax B

ax bx c

++ +

Therefore, if g(x) is a proper fraction f (x)

g(x) and can be resolved into real factors, then f (x)

g(x)

can be written in the following form :

Factor in the denominator Corresponding parital fraction

ax + bA

ax b+

(ax + b)22

A B

ax b (ax b)+

+ +

(ax + b)32 3

A B C


+ + +

ax2 + bx + c 2

Ax B

ax bx c

++ +

(ax2 + bx + c)22 2 2

Ax B Cx D

ax bx c (ax bx c)

+ +++ + + +

where A, B, C, D are arbitraty constants.

The rational functions which we shall consider for integration will be those whose denominatorscan be fracted into linear and quadratic factors.


2

2x 5dx

x x 2

+− −∫

392 MATHEMATICS

MODULE - VCalculus

Notes

Integration

Solution : 2

2x 5 2x 5

(x 2) (x 1)x x 2

+ +=− +− −

Let2x 5 A B

(x 2) (x 1) x 2 x 1

+ = +− + − + (1)

Multiplying both sides by (x - 2) (x + 1), we have

2x + 5 = A(x + 1) + B(x - 2)

Putting x = 2, we 9 = 3A or A = 3

Putting x = −1, we get 3 = −3B or B = −1

Substituting these values in (1), we have

2x 5 3 1

(x 2) (x 1) x 2 x 1

+ = −− + − +

⇒2x 5 3 1

dx dx dx(x 2) (x 1) x 2 x 1

+ = −− + − +∫ ∫ ∫

= 3log |x − 2| − log |x + 1| + C


(a)2

3 2

x x 1dx

x x 6x

− −− −∫ (b) 2

1dx

(x 1)(x 1)− +∫Solution :

(a)2 2

3 2

x x 1 x x 1

x(x 1) (x 2)x x 6x

− − − −=− +− −

Let2x x 1 A B C

x(x 1) (x 2) x x 3 x 2

− − = + +− + − + (1)

Multiplying both sides by x(x − 3) (x + 2), we have

x2 − x − 1 = A(x − 3) (x + 2) + Bx(x + 2) + Cx(x − 3)

Putting x = 3, we get 15B = 5 or B = 1

3

Putting x = 0, we get -6A = -1 or A = 1

6

Putting x = -2, we get 10C = 5 or C = 1

2Substituting these values in (1), we have

2x x 1 1 1 1

x(x 3)(x 2) 6x 3(x 3) 2(x 2)

− − = + +− + + +

MATHEMATICS 393

MODULE - VCalculus

Notes

Integration

⇒2

3 2

x x 1 1 1 1dx dx dx dx

6x 3(x 3) 2(x 2)x x 6x

− − = + +− +− −∫ ∫ ∫ ∫

1 1 1log | x | log | x 3 | log | x 2 | C

6 3 2= + − + + +

(b) 2 2

1 1 1

(x 1)(x 1)(x 1)(x 1)(x 1) (x 1)(x 1)= =

+ − +− + − +

Let 2 2

1 A B C

x 1 x 1(x 1)(x 1) (x 1)= + +

− +− + +

Multiplying both sides by (x2 − 1)(x + 1), we have

1 = A(x + 1)2 + B(x − 1) (x + 1) + C(x − 1)

Putting x = 1, we get1

A4

=

Putting x = −1, we get1

C2

= −

0 = A + B

⇒ B =1

4−

∴2 2

1 1 1 1 1 1dx dx dx dx

4(x 1) 4 x 1 2(x 1)(x 1) (x 1)= − −

− +− + +∫ ∫ ∫ ∫

1 1 1 1

log | x 1| log | x 1| C4 4 2 x 1

= − − + − − + +

1 1 1log | x 1| log | x 1| C

4 4 2(x 1)= − − + + +

+ Example 26.49 Evaluate :

3

2

x x 1dx

x 1

+ +−∫

Solution : Let I = 3

2

x x 1dx

x 1

+ +−∫

Now3

2 2

x x 1 2x 1 2x 1x x

(x 1)(x 1)x 1 x 1

+ + + += + = ++ −− −

∴2x 1

I x dx(x 1)(x 1)

+= + + − ∫

394 MATHEMATICS

MODULE - VCalculus

Notes

Integration

Let2x 1 A B

(x 1)(x 1) x 1 x 1

+ = ++ − + − (2)

⇒ 2x + 1 = A(x − 1) + B(x + 1)

Putting x = 1, we get B = 3

2

Putting x = −1, we get A = 1

2

Substituting the values of A and B in (2) and integrating, we have

2

2x 1 1 1 3 1dx dx dx

2 (x 1) 2 x 1(x 1)

+ = ++ −−∫ ∫ ∫

1 3log | x 1| log | x 1|

2 2= + + − (3)

∴ From (1) and (3), we have

2x 1 3I log | x 1| log | x 1| C

2 2 2= + + + − +


(a) 2

8dx

(x 2)(x 4)+ +∫ (b) 3

1dx

x(x 1)+∫

Solution :

(a) 2 2

8 A Bx C

x 2(x 2)(x 4) x 4

+= +++ + +

(As x2 + 4 is not factorisable into linear factors)

Multiplying both sides by (x + 2) (x2 + 4), we have

8 = A(x2 + 4) + (Bx + C) (x + 2)

On comparing the corresponding coefficients of powers of x on both sides, we get

0 A B

0 2B C A 1, B 1, C 2

8 4A 2C

= + = + ⇒ = = − == +

∴ 2 2

8 1 x 2dx dx dx

x 2(x 2)(x 4) x 4

−= −++ + −∫ ∫ ∫

2 4

1 1 2x dxdx dx 2

x 2 2 x 4 x 4= − +

+ + +∫ ∫ ∫

MATHEMATICS 395

MODULE - VCalculus

Notes

Integration

2 11 1 xlog | x 2 | log | x 4 | 2. tan C

2 2 2−= + − + + +

2 11 xlog | x 2 | log | x 4 | tan C

2 2−= + − + + +

(b) 3 2 2

1 1 A B Cx D

x x 1x(x 1) x(x 1)(x x 1) x x 1

+= = + +++ + − + − +

Multiplying both sides by x(x3 + 1) or x(x + 1) (x2 − x + 1), we have

I = A(x + 1)(x2 − x + 1) + Bx(x2 − x + 1) + (Cx + D) (x) (x + 1)

On comparing the corresponding coefficients of powers of x on both sides, we get

0 A B C

0 B C D 1 2 1A 1, B , C , D

0 B D 3 3 3

1 A

= + + = − + + ⇒ = = − = − == + =

3 2

1 1 1 2x 1

x 3(x 1)x(x 1) 3(x x 1)

−= − −++ − +

3 2

1 1 1 1 1 2x 1dx dx dx dx

x 3 x 1 3x(x 1) x x 1

−= − −++ − +∫ ∫ ∫ ∫

21 1log | x | log | x 1| log | x x 1| C

3 3= − + − − + +


dx

sin x sin 2x−∫Solution :

Let I = dx dx

sin x sin 2x sin x 2sin x cos x=

− −∫ ∫

dx

sin x(1 2cos x)=

−∫Multiplying numerator and denominator by sin x, we have

2 2

sin x dx sin x dxI

sin x(1 2cos x) (1 cos x)(1 2cos x)= =

− − −∫ ∫

sin x dx

(1 cos x)(1 cos x)(1 2cos x)=

+ − −∫

396 MATHEMATICS

MODULE - VCalculus

Notes

Integration

Put cos x = t, then − sinx dx − dt

∴dt

I(1 t)(1 t)(1 2t)

−=+ − −∫

Let1 A B C

(1 t)(1 t)(1 2t) 1 t 1 t 1 2t

− = + ++ − − − + −

Then, −1 = A(1 + t) (1 − 2t) + B(1 − t) (1 − 2t) + C(1 − t) (1 + t)

Putting t = −1, we get B = 1

6−

Putting t = 1, we get A = 1

2

Putting t = 1

,2

we get C = 4

3−

∴1 1 1 1 4 dt

I dt dt2 (1 t) 6 1 t 3 1 2t

= − −− + −∫ ∫ ∫

1 1 4log |1 t | log |1 t | log |1 2t | C

2 6 6= − − − + + − +

1 1 2log |1 cos x | log |1 cos x | log |1 2cos x | C

2 6 3= − − − + + − +


2

2sin 2 cosd

4 cos 4sin

θ − θ θ− θ − θ∫

Solution :

Let 2

2sin 2 cosI d

4 cos 4sin

θ − θ= θ− θ − θ∫

2

(4sin 1)cos d

2 sin 4sin

θ − θ θ=+ θ − θ∫

Let sin θ = t, then cos θ dθ = dt

∴ 2

4t 1I dt

3 t 4t

−=+ −∫

Let 2

4t 1 A B

t 3 t 13 t 4t

− = +− −+ −

Thus 4t − 1 = A(t − 1) + B(t − 3)

Put t = 1 then B = 3

2−

MATHEMATICS 397

MODULE - VCalculus

Notes

Integration

Put t = 1 then A = 11

2

∴11 1 3 dt

I dt2 t 3 2 t 1

= − − − ∫ ∫11 3

log | t 3 | log | t 1| C2 2

= − − − +

11 3log | sin 3 | log | sain 1| C

2 2= θ − − θ − +


3

3

tan tan

1 tan

θ + θ+ θ∫

Solution :

Let3 2

3 3

tan tan tan (1 tan )I d d

1 tan 1 tan

θ + θ θ + θ= θ = θ+ θ + θ∫ ∫

2

3

tan secd

1 tan

θ + θ= θ+ θ∫

Let tan θ = t, then sec2θ dθ = dt

∴ 3 2

t dt t dtI

1 t (1 t)(1 t t )= =

+ + − +∫ ∫ (1)

Let 2 2

t dt A Bt C

1 t(1 t)(1 t t ) 1 t t

+= +++ − + − +

Then t = A(1 − t + t2) + (Bt + C) (1 + t)

Comparing the coefficients of t, we get

A + B = 0, −A + B + C = 1, A + C = 0

⇒1 1 1

A , B , C3 3 3

= − = =

∴ 2

1 dt 1 t 1I dt

3 1 t 3 1 t t

+= − ++ − +∫ ∫

2

1 dt 1 2t 2dt

3 1 t 6 t t 1

+= − ++ − +∫ ∫

2

1 dt 1 (2t 1) 3dt

3 1 t 6 t t 1

− += − ++ − +∫ ∫

398 MATHEMATICS

MODULE - VCalculus

Notes

Integration

2 22

1 dt 1 (2t 1)dt 1 1

3 1 t 6 2t t 1 1 3t

2 2

−= − + ++ − + − +

∫ ∫ ∫

2 1

1t1 1 1 2 2log |1 t | log | t t 1| . tan

3 6 2 3 32

−

−

= − + + − + +

2 11 1 1 2t 1log |1 t | log | t t 1| tan C

3 6 3 3− − = − + + − + + +

2 11 1 1 2 tan 1log |1 tan | log | tan tan 1| tan C

3 6 3 3− θ − = − + θ + θ − θ + + +


Evaluate the following :

1. (a) 24x 5 dx−∫ (b) 2x 3x dx+∫(c) 23 2x 2x dx− −∫

2. (a)x 1

dx(x 2)(x 3)

+− −∫ (b) 2

xdx

x 16−∫

3. (a)3

2

xdx

x 4−∫ (b)2

2

2x x 1dx

(x 1) (x 2)

+ +− +∫

4.2

3

x x 1dx

(x 1)

+ +−∫

5. (a)sin x

dxsin 4x∫ (b)

1 cos xdx

cos x(1 cos x)

−+∫

LET US SUM UP

● Integration is inverse of differention

● Standard form of some indefinite integrals

(a) nx dx∫n 1x

C (n 1)n 1

+= + ≠ −

+

MATHEMATICS 399

MODULE - VCalculus

Notes

Integration

(b)1

dxx∫ = log |x| + C

(c) sin x ax∫ = −cos x + C

(d) cos x dx∫ = sin x + C

(e) 2sec x dx∫ = tan x + C

(f) 2cosec x dx∫ = −cot x + C

(g) sec x tan x dx∫ = sec x + C

(h) cosec x cot x dx∫ = −cosec x + C

(i) 2

1dx

1 x−∫ = sin−1 x + C

(j) 2

1dx

1 x+∫ = tan−1 x + C

(k) 2

1dx

x x 1−∫ = sec−1 x + C

(l) xe dx∫ = ex + C

(m) xa dx∫xa

C (a 0 and a 1)log a

= + > ≠

● Properties of indefinite integrals

(a) [f (x) g(x)]dx±∫ f (x)dx g(x)dx= ±∫ ∫(b) kf (x)dx∫ k f (x)dx= ∫

(i) n(ax b) dx+∫n 11 (ax b)

C (n 1)a n 1

++= + ≠ −+

(ii)1

dxax b+∫

1log | ax b | C

a= + +

(iii) sin(ax b) dx+∫1

cos(ax b) Ca

−= + +

(iv) cos(ax b) dx+∫1

sin(ax b) Ca

= + +

(v) 2sec (ax b) dx+∫1

tan(ax b) Ca

= + +

400 MATHEMATICS

MODULE - VCalculus

Notes

Integration

(vi) 2cosec (ax b)dx+∫1

cot(ax b) Ca

= − + +

(vii) sec (ax b) tan(ax b) dx+ +∫1

sec (ax b) Ca

= + +

(viii) cosec (ax b) cot (ax b)dx+ +∫1

cosec (ax b) Ca

= − + +

(xi)ax b ax b1

e dx e Ca

+ += +∫f (x)

dx log | f (x) | Cf (x)

′= +∫

● (i) tan x dx log | cos x | C log | sec x | C= − + = +∫(ii) cot x dx log | sin x | C= +∫(iii) sec x dx log | sec x tan x | C= + +∫(iv) cosec x dx log | cosecx cot x | C= − +∫

(i) 2 2

1 1 a xdx log C

2 a xa x

+= +−−∫ (ii) 2 2

1 1 x adx log C

2a x ax a

−= ++−∫

(iii)1

2 2

1 1 xdx tan C

a ax a−= +

+∫ (iv)1

2 2

1 xdx sin C

aa x

−= +−

∫

(v)2 2

2 2

1log x x a C

x a= + − +

+∫

(vi)2 2

2 2

dxlog x x a C

x a= + + +

+∫

● (i)2

14

1xx 1 1 xdx tan C

x 1 2 2−

− + = + +

∫

(ii)2

4

1x 2x 1 1 xdx log C

1x 1 2 2 x 2x

+ −− = ++ + +

∫

(iii)2

14

1 1x x 2x 1 2 xdx tan log C

1x 1 2 2 2 x 2x

−

− + − = + + + + +

∫

MATHEMATICS 401

MODULE - VCalculus

Notes

Integration

(iv) 14

1 1x x 21 1 12 xdx tan log C

1x 1 2 2 2 2 2 x 2x

−

− + − = − + + + +

∫

● (i)2

14 2

1xx 1 1 xdx tan C

x x 1 3 3−

− + = + + +

∫

(ii)2

4 2

1x 1x 1 1 xdx log C

12x x 1 x 1x

+ − − = + + + + +

∫

(iii)1 1

4 2

1 1 1 tan x 1 1 cot x 1dx tan tan C

2x x 1 2 2 2 2− − − − = + + + +

∫

(iv) ( ) 1tan x cot x dx 2 sin (sin x cos x) C−+ = − +∫● Integral of the product of two functions

I function × II function − [ ]Derivative of I function Integral of II function dx×∫● [ ]x xe f (x) f (x) dx e f (x) C′+ = +∫

●2 2 2 2 2 11 x

a x dx x x a a sin C2 a

− − = − + + ∫

2 2 22 2 2 2x x a a

x a dx log x x a C2 2

−− = − + − +∫

2 2 22 2 2 2x a x a

a x dx log x a x C2 2

++ = + + + +∫● Rational fractions are of following two types :

(i) Proper, where degree of variable of numerator < denominator.

(ii) Improper, where degree of variable of numerator > denominator.

● If g(x) is a proper fraction f (x)

g(x) can be resolved into real factors, then f (x)

g(x) can be

written in the following form :

402 MATHEMATICS

MODULE - VCalculus

Notes

Integration

Factors in denominator Corresponding partial fraction

ax + bA

ax b+

(ax + b)22

A B

ax b (ax b)+

+ +

(ax + b)32 3

A B C


+ + +

ax2 + bx + C 2

Ax B

ax bx c

++ +

(ax2 + bx + c)22 2 2

Ax B Cx D

ax bx c (ax bx c)

+ +++ + + +

where A, B, C, D are arbitrary constants.

SUPPORTIVE WEBSITES


● http://mathworld.wolfram.com

TERMINAL EXERCISE

Integrate the following functions w.r.t. x :

1.3 3

2 2

sin x cos x

sin x cos x

+2. 1 sin 2x+

3. 2 2

cos 2x

cos x sin x4. (tan x − cot x)2

5. 2 2

4 1

1 x 1 x−

+ −6.

22sin x

1 cos 2x+

7.22cos x

1 cos 2x−8.

2x x

sin cos2 2

+

9.2

x xcos sin

2 2 −

10. cos(7x − π)

11. sin (3x + 4) 12. sec2(2x + b)

MATHEMATICS 403

MODULE - VCalculus

Notes

Integration

13.dx

sin x cos x−∫ 14. 2 1

1dx

(1 x ) tan x−+∫

15.cosec x

dxx

log tan2

∫ 16.cot x

dx3 4log sin x+∫

17.dx

sin 2x log tan x∫ 18.x

x

e 1dx

e 1

+−∫

19. 4sec x tan x dx∫ 20. x xe sin e dx∫

21. 2

x dx

2x 3+∫ 22.

2sec xdx

tan x∫

23. 225 9x dx−∫ 24. 22ax x dx−∫25. 23x 4 dx+∫ 26. 21 9x dx+∫

27.2

2 2

x dx

x a−∫ 28. 2 2

dx

sin x 4cos x+∫

29.dx

2 cos x+∫ 30. 2

dx

x 6x 13− +∫

31. 2

dx

1 3sin x+∫ 32.2

2 2

xdx

x a−∫

33. 2

dx

1 4cos x−∫ 34.sin x

dxsin 3x∫

35. 2

dx

1 4cos x−∫ 36. 2sec (ax b)dx+∫

37.dx

x(2 log x)+∫ 38.5

6

xdx

1 x+∫

39.cos x sin x

dxsin x cos x

−+∫ 40.

cot xdx

log sin x∫

41.2sec x

dxa b tan x+∫ 42.

sin xdx

1 cos x+∫43. 2cos x dx∫ 44. 3sin x dx∫45. sin 5x sin 3x dx∫ 46. 2 3sin x cos x dx∫47. 4sin x dx∫ 48.

1dx

1 sin x+∫

404 MATHEMATICS

MODULE - VCalculus

Notes

Integration

49. 3tan x dx∫ 50.cos x sin x

dx1 sin 2x

−+∫

51.2cosec x

dx1 cot x+∫ 52. 2

1 x cos 2xdx

x sin 2x 2x

+ ++ +∫

53.sec cosec d

log tan

θ θ θθ∫ 54.

cot d

log sin

θ θθ∫

55. 2

dx

1 4x+∫ 56.1 tan

d1 tan

− θ θ+ θ∫

57.1x

2

1e dx

x

−

∫ 58. 2 2 2 2

sin x cos x dx

a sin x b cos x+∫

59.dx

sin x cos x+∫ 60.x 1

2

1e cos x dx

1 x

− − −

∫

61.x

2

sin x cos xe dx

cos x

+ ∫ 62.

1 1 cos xtan dx

1 cos x− −

+∫

63. 1 1 xcos 2cot dx

1 x− −

+ ∫ 64.

( )

1

32 2

sin xdx

1 x

−

−∫

65. x log x dx∫ 66. x xe (1 x) log(xe ) dx+∫

67. 2

log xdx

(1 x)+∫ 68. x xe (1 x) log(xe ) dx+∫69. cos(log x) dx∫ 70. log(x 1) dx+∫

71.2

2

x 1dx

(x 1) (x 3)

+− +∫ 72. 2

sin cosd

cos cos 2

θ θ θθ − θ −∫

73. 5

dx

x(x 1)+∫ 74.2

2 2

x 1dx

(x 2)(2x 1)

++ +∫

75.log x

dxx(1 log x)(2 log x)+ +∫ 76. x

dx

1 e−∫

MATHEMATICS 405

MODULE - VCalculus

Notes

Integration

ANSWERS


1.2 7 7 7 77 2 2 2 22 2 2 2 2

x 1, x 2, x 3, x 4, x 57 7 7 7 7

+ + + + +

2. (a)6x

C6

+ (b) sin x + C (c) 0

3. (a)7x

C7

+ (b) 6

1C

6x+ (c) log |x| + C

(d)

x35

C3

log5

+

(e)433

x C4

+ (f) 8

1C

8x

− +

(g) 2 x C+ (h)1

99x C+4. (a) −cosec θ + C (b) sec θ + C

(c) tan θ + C (d) − cot θ + C


1. (a)2x 1

x C2 2

+ + (b) −x + tan−1x + C

(c)3

10 22x x 2 x C

3− + + (d) 5 4 3

1 3 2 78x C

xx 4x 3x− − + + − +

(e)3

1xx tan x C

3−− − + (f)

2x4x 4log x C

2+ + +

2. (a)1

tan x C2

+ (b) tan x − x + C

(c) −2 cosec x + C (d)1

cot x C2

− +

(e) − sec x + C (f) − cot x + cosec x + C

3. (a) 2 sin x C+ (b) 2 cos x C− +

(c)1

cot x C2

− +

4. (a)322

(x 2) C3

+ + (b)16

17 1C

16 (x 1)

− ++

406 MATHEMATICS

MODULE - VCalculus

Notes

Integration


1. (a)1

cos(4 5x) C5

− + (b)1

tan(2 3x) C3

+ +

(c) log sec x tan x C4 4

π π + + + +

(d)1

sin(4x 5) C4

+ +

(e)1

sec (3x 5) C3

+ + (f)1

cosec (3 5x) C5

− + +

2. (a) 3

1C

12(3 4x)+

− (b)51

(x 1) C5

+ +

(c)111

(4 7x) C77

− − + (d)41

(4x 5) C16

− +

(e)1

log | 3x 5 | C3

− + (f)2

5 9x C9

− − +

(g)31

(2x 1) C6

+ + (h) log | x 1| C+ +

3. (a)2x 11

e C2

+ + (b)3 8x1

e C8

−− +

(c) (7 4x)

1C

4e + +

4. (a)1 sin 2x

x C2 2

+ + (b)

1 3 1cos 2x cos 6x C

32 2 6 − + +

(c)1 cos7x

cos x C2 7

− − + (d)

1 sin 6x sin 2xC

2 6 2 + +


1. (a)21

log | 3x 2 | C6

− + (b) log |x2 + x + 1| + C

(c) log |x2 + 9x + 30| + C (d)31

log | x 3x 3 | C3

+ + +

(e) log |x2 + x − 5| + C (f) 2 log | 5 x | C+ +(g) log |8 + log x | + C

2. (a)x1

log | a be | Cb

+ + (b) tan−1 (ex) + C

MATHEMATICS 407

MODULE - VCalculus

Notes

Integration


1. (a)3 x 3

x log C2 x 3

−+ ++ (b) ( )1 xtan e C− +

(c)1 21

tan (x ) C2

− + (d)11 3x

sin C3 4

− +

(e)11

tan (2 tan x) C2

− + (f)1 x 1

sin C2

− + +

(g) 11 x 1tan C

3 6 6− + +

(h)1 x 2

sin C3

− + +

(i) 11 xsec C

22 3− + (j)

211 tan 1

tan C2 2 tan

− θ − + θ

(k) x 2xlog e 1 e C+ + + (l) 1 2sin x 1 x C− − − +

(m)1 x a

sin Ca

− − + (n) 1 31 4

sin x C4 3

− +

(o) 2 2x 1 log x x 1 C+ + − + + (p)21 2x 9 4x

log C2 2

+ + +

(q) 21log 2cos 4cos 1 C

2− θ + θ − +

(r) 2log tan x tan x 4 C+ − +

(s) 1 x 2tan C

1− + +

CHECK YOUR PROGRESS 26.61. (a) − x cos x + sin x + C

(b) ( )21 x cos 2x sin 2x1 x sin 2x C

2 2 4+ + − +

(c)x cos 2x 1 sin 2x

C2 2 2

− + +

2. (a) x tan x − log |sec x| − x + C

(b) 3 21 1 1 1x x sin 2x x cos 2x sin 2x C

6 4 4 8− − + +

408 MATHEMATICS

MODULE - VCalculus

Notes

Integration

3. (a)4 4x log 2x x

C4 16

− + (b)3 3x x

x log x x C3 9

− − + +

(c) x(log x)2 − 2x log x + 2x + C

4. (a)1 n 1 nx x

log x C1 n (1 n)

− −

2− +− −

(b) log x . [log (log x) − 1] + C

5. (a)2

3x x 2x 2e C

3 9 27

− + +

(b)

4x 4xe ex C

4 16− +

6.2

2x 1(log x) log x C

2 2 − + +

7. (a)1 2x sec x log x x 1 C− − + − +

(b)2

1 1x x 1cot x cot x C

2 2 2− −+ + +

CHECK YOUR PROGRESS 26.71. (a) ex sec x + C (b) ex log|sec x + tan x| + C

2. (a)x1

e Cx

+ (b) ex sin−1x + C

3.x

2

eC

(1 x)+

+ 4.xe

C1 x

++

5.x

x tan C2

+

6.x1

e (sin 2x 2cos 2x) C5

− +


1. (a)2 25 5 5

x x log x x C4 4 4

− − + − +

(b)2 2(2x 3) 9 3

x 3x log x x 3x C4 8 2

+ + − + + + +

(c)2 11 7 2x 1

(2x 1) 3 2x 2x sin C4 4 2 7

− + + − − + +

MATHEMATICS 409

MODULE - VCalculus

Notes

Integration

2. (a) 4 log | x 3 | 3log | x 2 | C− − − +

(b)1

log | x 4 | log | x 4 | C2

− + + +

3. (a)2x

2[log | x 2 | log | x 2 |] C2

− − + + +

(b)11 7 4

log | x 1| log (x 2) C9 9 3(x 1)

− + + − +−

4. 2

3 3log | x 1| C

(x 1) 2(x 1)− − − +

− −

5. (a)1 1

log|1 sin x | |1 sin x |8 8

− − +

1 1log 1 2 sin x log 1 2 sin x C

4 2 4 2− − + + +

(b)x

log| sec x tan x | 2 tan C2

+ − +

TERMINAL EXERCISE1. sec x − cosec x + C 2. sin x − cos x + C

3. − cot x − tan x + C 4. tan x − cot x − 4x + C

5. 4 tan−1x − sin−1 x + C 6. tan x − x + C

7. − cot x − x + C 8. x − cos x + C

9. x + cos x + C 10.sin(7x )

C7

− π +

11.cos(3x 4)

C3

− + + 12.tan(2x b)

C2

+ +

13.1

log cosec x cot x C4 42

π π − − − +

14. 1log | tan x | C− + 15.x

log log tan C2

+

16.1

log 3 4log sin x C4

+ + 17.1

log log tan x C2

+

18.x x2 22 log | e e | C

−

− + 19.41

sec x C4

+

410 MATHEMATICS

MODULE - VCalculus

Notes

Integration

20. −cos cx + C 21.22x 3

C2

+ +

22. 2 tan x C+

23. ( )2 11 25 3x 25 9x sin x C

6 6 5− − + +

24. 2 2 11 1 x a(x a) 2ax x a sin C

2 2 a− − − − + +

25.2 2x 3x 4 2 3x x 4

log C2 23

+ + ++ +

26.2

2x 9x 1 1log 3x 1 9x C

2 6

+ + + + +

27.2 2 2 2 21 1

x x a a log x x a C2 2

− + + − +

28.11 tan x

tan C2 2

− + 29. 1

xtan

2 2tan C

3 3−

+

30.11 x 3

tan C2 2

− − + 31.

11tan (2 tan x) C

2− +

32.a x a

x log C2 x a

−+ ++ 33.

4

4

1 9 x 3log C

12 9 x 3

+ − ++ +

34.1 3 tan x

log C2 3 3 tan x

+ +−

35.1 tan x 2

log C2 2 tan x 2

− ++

36.1

tan(ax b) Ca

+ + 37. log |(2 + log x)| + C

38.61

log(1 x ) C6

+ + 39. log |sinx + cos x| + C

40. log |log (sin x)| + C 41.1

log a b tan x Cb

+ +

42. −log |1 + cos x| + C 43.1 sin 2x 1

x C2 2 2

+ +

MATHEMATICS 411

MODULE - VCalculus

Notes

Integration

44.3cos x

cos x C3

− + + 45.1 sin 2x 1 sin 8x

C2 2 2 8

− +

46.5

31 sin xsin x C

3 5− + 47.

1[12x 8sin 2x sin 4x] C

32− + +

48. tan x − sec x + C 49. 2tan x

log | cos x | C2

+ +

50.1

Ccos x sin x

− ++ 51.

1log C

1 cot x+

+

52. 21log x sin 2x 2x C

2+ + + 53. log |tan θ| + C

54. log |log sin θ| + C 55.11

tan 2x2

−

56. log |cos θ + sin θ| + C 57.1xe C

−+

58. 2 2 2 22 2

1log a sin x b cos x C

2(a b )+ +

−

59.1

log sec x tan x C4 42

π π − + − +

60. ex cos−1x + C 61. ex sec x + C

62.21

x C4

+ 63.21

x C2

− +

64.1

2

2

x sin x 1log|1 x | C

21 x

−+ − +

−65.

3

22 2x log x C

3 3 − +

66. ( )x xxe log xe 1 C − +

67.1

log | x | log | x | log | x 1| C1 x

− + − + ++

68.x

x1 ee (2sin 2x cos 2x) C

2 10− + +

69. [ ]xcos(log x) sin(log x) C

2+ +

70. x log | x 1| x log | x 1| C+ − + + +

71.3 1 5

log | x 1| log | x 3 | C8 2(x 1) 8

− − + + +−

412 MATHEMATICS

MODULE - VCalculus

Notes

Integration

72.2 1

log cos 2 log cos 1 C3 3

− θ − − θ + +

73.5

5

1 xlog C

5 x 1+

+

74. ( )1 11 xtan tan 2x C

3 2 2− − + +

75.2(2 log x)

log C1 log x

+ ++

76.x

x

elog C

1 e+

−

412/1

Integration

26.9 Reduction Formulae

There are many functions whose integrals cannot be reduced to one or the other of the well

known standard forms of integration. However, in some cases these integrals can be connected

algebraically with integrals of other expressions which are directly integrable or which may be easier to

integrate than the original functions. Such connecting algebraic relations are called reduction formulae.

The formulae connect an integral with another which is of the same type, but is of lower degreee or

order or at any rate relatively easier to integrate. In this section, we illustrate the method of integration

by successive reduction.

26.9.1 Reduction formula for ∫ n axx e dx n being a positive integer

Let I n axn x e dx= ∫

On using formula for integration by parts, we get

1In ax ax

nn

x e en x dx

a a−= − ∫

1n ax

n axx e nx e dx

a a−= − ∫

1In ax

nx e n

a a −= −

This is called a reduction formula for n axx e dx∫ . Now In−1 in turn can be connected to In−2.

By successive reduction of n, the original integral In finally depends on I0 where

I0 = ax

ax ee dx

a=∫

412/2

Example 1

3 5xx e dx∫Sol: We take a = 5 and use the reduction formula for n = 3, 2, 1 in that order.

Then we have

I3 = 3 5

3 52

3I

5 5

xx x e

x e dx= −∫

I2

2 5

12

I5 5

xx e= −

I1

5

01

I5 5

xxe= −

and I0

5

5

xec= +

Hence I3

3 52 5 5 5

2 3 4

3 6 6.

5 5 5 5

xx x xx e

x e x e e c= − + − +

1. Theorem : Reduction formula for sinn x dx∫ for an integer n> 2

Proof: Let In = sinn x dx∫In = 1sin sinn x x dx−∫

1sin ( cos )n dx x dx

dx−= −∫

1 2sin ( cos ) ( 1) sin cos ( cos )n nx x n x x x dx− −= − − − −∫1 2 2sin cos ( 1) sin (1 sin )n nx x n x x dx− −= − + − −∫

1 2sin cos ( 1) sin ( 1) sinn n nx x n x dx n x dx− −= − + − − −∫ ∫1

2sin cos ( 1) I ( 1) Inn nx x n n−−= − + − − −

Hence1

2sin cos 1

I In

n nx x n

n n

−

−− −

= +

412/3

This is called reduction formula for sinn x dx∫If n is even, after succesive reduction, we get

00 1I (sin )x dx x c= = +∫

If n is odd, after successive reduction, we get

11 2I (sin ) cosx dx x c= = − +∫

Example 2

Evaluate 4sin x dx∫ .

Sol: On using the reduction formula for 4sin x dx∫ with n = 4 and 2 in that order we have

44I sin x dx= ∫

3

2sin cos 3

I4 4

x x= − +

3

0

sin cos 3 sin cos 1+ I

4 4 2 2

x x x x = − + − 3sin cos 3 3

sin cos .4 8 8

x xx x x c= − − + +

Notes :

(1) Reduction formula for Tannx dx∫ for an integer n> 2.

Let1

2Tan

I Tan I1

nn

n nx

xn

−

−= = −−∫

where n is even, In will finally depend on

0 1I dx x c= = +∫when n is odd, In will finally depend on

1 2I Tan log | sec |x dx x c= = +∫ (2) Reduction formula for secnx dx∫ for an integer n> 2

Let2

21 2

I sec sec Tan I .1 1

n nn n

nx dx x x

n n−

−−= = +

− −∫

412/4

when n is even the last integral to which In can be reduced is 0 1I dx x c= = +∫When n is odd, the ultimate integral is I1.

which is 1 2I sec log | sec tan |x dx x x c= = + +∫Example 3

5sec x dx∫Sol : On using n = 5 in the above reduction formula

35

5 3sec tan 3

I sec I4 4x x

x dx= = +∫3

1sec tan 3 sec Tan 3

I4 4 2 8x x x x= + +

3sec tan 3 3sec Tan

4 8 8

x xx x= + + log | Sec x + Tan x | + c.


I. Evaluate the following integrals.

1. 2 3xx e dx−∫2. 3 axx e dx∫

II.Evaluate the following integrals.

1. 4Tan x dx∫2. 4cos x dx∫3. 3sec x dx∫

III 1. If In = cosnx dx∫ . Then show that

12

1 1I cos sin Inn n

nx x

n n−

−−= +

412/5

Let Us Sum Up

1. If In = ,n axx e dx∫ Then

In = 1In ax

nx e n

a a −− For a positive inter n.

2. If In = sinn x dx∫ then

In

1

2sin cos 1

In

nx x n

n n

−

−− −= + , for an inter n > 2.

3. If In = cosn x dx∫ , Then

In

1

2cos sin 1

In

nx x n

n n

−

−−= + , for an integer n > 2

4. If In = Tannx dx∫ , Then

1

2Tan

I I1

n

n nx

n

−

−= −−

, for an integer n> 2.

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Answers


I (1)3

2(9 6 2)27

xex x c

−− + + +

(2) 3 3 2 24

( 3 6 6)axe

a x a x ax ca

− + − +

II(1)3sin cos 3 3

sin cos4 8 8x x

x x x c− − + +

(2) 31 3 3cos sin cos sin

4 8 8x x x x x c+ + +

MATHEMATICS 413

Notes

MODULE - VCalculus

Definite Integrals

27

DEFINITE INTEGRALS

In the previous lesson we have discussed the anti-derivative, i.e., integration of a function.Thevery word integration means to have some sort of summation or combining of results.

Now the question arises : Why do we study this branch of Mathematics? In fact the integrationhelps to find the areas under various laminas when we have definite limits of it. Further we willsee that this branch finds applications in a variety of other problems in Statistics, Physics, Biology,Commerce and many more.

In this lesson, we will define and interpret definite integrals geometrically, evaluate definite integralsusing properties and apply definite integrals to find area of a bounded region.


define and interpret geometrically the definite integral as a limit of sum;

evaluate a given definite integral using above definition;

state fundamental theorem of integral calculus;

state and use the following properties for evaluating definite integrals :

(i)b a

a bf x dx f x dx (ii)

c b c

a a bf x dx f x dx f x dx

(iii)2a a a

0 0 0f x dx f x dx f 2a x dx

(iv)b b

a af x dx f a b x dx

(v)a a

0 0f x dx f a x dx

MATHEMATICS

Notes

MODULE - VCalculus

414

Definite Integrals

(vi)2a a

0 0f x dx 2 f x dx if f 2a x f x

0 if f 2a x f x

(vii)a a

a 0f x dx 2 f x dx if f is an even function of x

= 0 if f is an odd function of x.

apply definite integrals to find the area of a bounded region.

EXPECTED BACKGROUND KNOWLEDGEKnowledge of integration

Area of a bounded region

27.1 DEFINITE INTEGRAL AS A LIMIT OF SUM

In this section we shall discuss the problem of finding the areas of regions whose boundary isnot familiar to us. (See Fig. 27.1)

Fig. 27.1 Fig. 27.2

Let us restrict our attention to finding the areas of such regions where the boundary is notfamiliar to us is on one side of x-axis only as in Fig. 27.2.

This is because we expect that it is possible to divide any region into a few subregions of thiskind, find the areas of these subregions and finally add up all these areas to get the area of thewhole region. (See Fig. 27.1)

Now, let f (x) be a continuous function defined on the closed interval [a, b]. For the present,assume that all the values taken by the function are non-negative, so that the graph of thefunction is a curve above the x-axis (See. Fig.27.3).

MATHEMATICS 415

Notes

MODULE - VCalculus

Definite Integrals

Fig. 27.3

Consider the region between this curve, the x-axis and the ordinates x = a and x = b, that is, theshaded region in Fig.27.3. Now the problem is to find the area of the shaded region.

In order to solve this problem, we consider three special cases of f (x) as rectangular region ,triangular region and trapezoidal region.

The area of these regions = base × average height

In general for any function f (x) on [a, b]

Area of the bounded region (shaded region in Fig. 27.3 ) = base × average height

The base is the length of the domain interval [a, b]. The height at any point x is the value of f (x)at that point. Therefore, the average height is the average of the values taken by f in [a, b]. (Thismay not be so easy to find because the height may not vary uniformly.) Our problem is how tofind the average value of f in [a,b].

27.1.1 Average Value of a Function in an IntervalIf there are only finite number of values of f in [ a,b], we can easily get the average value by theformula.

Average value of f in Sumof thevaluesof f in a ,ba , b

Numbersof values

But in our problem, there are infinite number of values taken by f in [ a, b]. How to find theaverage in such a case? The above formula does not help us, so we resort to estimate theaverage value of f in the following way:First Estimate : Take the value of f at 'a' only. The value of f at a is f (a). We take this value,namely f (a), as a rough estimate of the average value of f in [a,b].Average value of f in [a, b] ( first estimate ) = f (a) (i)Second Estimate : Divide [a, b] into two equal parts or sub-intervals.

Let the length of each sub-interval be h, b ah2

.

Take the values of f at the left end points of the sub-intervals. The values are f (a) and f (a + h)

MATHEMATICS

Notes

MODULE - VCalculus

416

Definite Integrals(Fig. 27.4)

Fig. 27.4

Take the average of these two values as the average of f in [a, b].

Average value of f in [a, b] (Second estimate)f a f a h b a, h

2 2(ii)

This estimate is expected to be a better estimate than the first.

Proceeding in a similar manner, divide the interval [a, b] into n subintervals of length h

(Fig. 27.5), b ah

n

Fig. 27.5

Take the values of f at the left end points of the n subintervals.

The values are f (a), f (a + h),......,f [a + (n-1) h]. Take the average of these n values of f in[a, b].Average value of f in [a, b] (nth estimate)

f a f a h .......... f a n 1 h,

nb ah

n(iii)

For larger values of n, (iii) is expected to be a better estimate of what we seek as the averagevalue of f in [a, b]

Thus, we get the following sequence of estimates for the average value of f in [a, b]:

MATHEMATICS 417

Notes

MODULE - VCalculus

Definite Integrals

f (a)1 f a f a h ,2

b ah2

1 f a f a h f a 2h ,3

b ah3

.........

.........1 f a f a h ........ f{a n 1 h} ,n

b ahn

As we go farther and farther along this sequence, we are going closer and closer to our destina-tion, namely, the average value taken by f in [a, b]. Therefore, it is reasonable to take the limit ofthese estimates as the average value taken by f in [a, b]. In other words,

Average value of f in [a, b]

n

1lim f a f a h f a 2h ...... f a n 1 h ,n

b ahn

(iv)

It can be proved that this limit exists for all continuous functions f on a closed interval [a, b].Now, we have the formula to find the area of the shaded region in Fig. 27.3, The base is

b a and the average height is given by (iv). The area of the region bounded by the curve f(x), x-axis, the ordinates x = a and x = b

n

1b a lim f a f a h f a 2h ...... f a n 1 h ,n

n 0

1 b alim f a f a h ........ f{a n 1 h} , h

n n(v)

We take the expression on R.H.S. of (v) as the definition of a definite integral. This integral isdenoted by

b

af x dx

read as integral of f (x) from a to b'. The numbers a and b in the symbolb

af x dx are called

respectively the lower and upper limits of integration, and f (x) is called the integrand.

Note : In obtaining the estimates of the average values of f in [a, b], we have taken the leftend points of the subintervals. Why left end points?

Why not right end points of the subintervals? We can as well take the right end points of the

MATHEMATICS

Notes

MODULE - VCalculus

418

Definite Integralssubintervals throughout and in that case we getb

na

1f x dx b a lim f a h f a 2h ...... f b ,n

b ahn

h 0lim h f a h f a 2h ...... f b (vi)

Example 27.1 Find2

1x dx as the limit of sum.

Solution : By definition,b

na

1f x dx b a lim f a f a h ........ f a n 1 h ,n

b ahn

Here a = 1, b = 2, f (x) = x and 1hn

.

2

n1

1 1 n 1xdx lim f 1 f 1 ........ f 1n n n

n

1 1 2 n 1lim 1 1 1 ........ 1

n n n n

n ntimes

1 1 2 n 1lim 1 1 ...... 1 ........n n n n��

n

1 1lim n 1 2 ...... n 1

n n

n

n 1 .n1lim n

n n.2

n 1 .nSince1 2 3 .... n 1

2

n

1 3n 1lim

n 2

n

3 1 3lim

2 2n 2

MATHEMATICS 419

Notes

MODULE - VCalculus

Definite Integrals

Example 27.2 Find 2

x

0e dx as limit of sum.

Solutions : By definitionb

h 0a

f x dx lim h f a f a h f a 2h ..... f a n 1 h

whereb ah

n

Here x 2 0 2a 0,b 2,f x e and hn n

2x

h 00e dx lim h f 0 f h f 2h ....... f n 1 h

n 1 h0 h 2hh 0lim h e e e ....... e

nh0

hh 0

e 1lim h e

e 1

n2 n 1 r 1

Since a ar ar ....... ar ar 1

nh 2

h hh 0 h 0

e 1 h e 1lim h limhe 1 e 1

h ( nh 2∵ )

2 2

hh 0

e 1 e 1lim1e 1

h

2e 1h

h 0

e 1lim 1

h∵

In examples 27.1 and 27.2 we observe that finding the definite integral as the limit of sum is quitedifficult. In order to overcome this difficulty we have the fundamental theorem of integral calculuswhich states thatTheorem 1 : If f is continuous in [ a, b] and F is an antiderivative of f in [a, b] then

b

af x dx F b F a ......(1)

The difference F (b) F (a) is commonly denoted by baF x so that (1) can be written as

bb ba a

af x dx F x or F x

MATHEMATICS

Notes

MODULE - VCalculus

420

Definite IntegralsIn words, the theorem tells us that

b

af x dx (Value of antiderivative at the upper limit b)

(Value of the same antiderivative at the lower limit a)

Example 27.3 Find 2

1x dx

Solution :22 2

11

xx dx2

4 1 32 2 2

Example 27.4 Evaluate the following

(a)2

0cosx dx (b)

22x

0e dx

Solution : We know that

cosxdx sinx c

220

0cosxdx s i n x

sin sin 0 1 0 12

(b)22 2x

2x

00

ee dx ,2

x xe dx e∵

4e 12

Theorem 2 : If f and g are continuous functions defined in [a, b] and c is a constant then,

(i)b b

a ac f x dx c f x dx

(ii)b b b

a a af x g x dx f x dx g x dx

(iii)b b b


MATHEMATICS 421

Notes

MODULE - VCalculus

Definite Integrals

Example 27.5 Evaluate 2

2

04x 5x 7 dx

Solution :2 2 2 2

2 2

0 0 0 04x 5x 7 dx 4x dx 5x dx 7 dx

2 2 22

0 0 04 x dx 5 xdx 7 1 dx

2 23 220

0 0

x x4. 5 7 x3 2

8 44. 5 7 2

3 2

32 10 143

443


1. Find5

0x 1 dx as the limit of sum. 2. Find

1x

1e dx as the limit of sum.

3. Evaluate (a)4

0sinx dx (b)

2

0sinx cosx dx

(c)1

20

1 dx1 x (d)

23 2

14x 5x 6x 9 dx

27.2 EVALUATION OF DEFINITE INTEGRAL BYSUBSTITUTION

The principal step in the evaluation of a definite integral is to find the related indefinite integral.In the preceding lesson we have discussed several methods for finding the indefinite integral.One of the important methods for finding indefinite integrals is the method of substitution. Whenwe use substitution method for evaluation the definite integrals, like

3

22

x dx1 x ,

2

20

s i n x dx,1 cos x

MATHEMATICS

Notes

MODULE - VCalculus

422

Definite Integralsthe steps could be as follows :(i) Make appropriate substitution to reduce the given integral to a known form to integrate.

Write the integral in terms of the new variable.(ii) Integrate the new integrand with respect to the new variable.(iii) Change the limits accordingly and find the difference of the values at the upper and lower

limits.Note : If we don't change the limit with respect to the new variable then after integratingresubstitute for the new variable and write the answer in original variable. Find the values ofthe answer thus obtained at the given limits of the integral.


22

x dx1 x

Solution : Let 21 x t

2x dx = dt or1xdx dt2

When x = 2, t = 5 and x =3, t = 10. Therefore, 5 and 10 are the limits when t is the variable.

Thus3 10

22 5

x 1 1dx dt2 t1 x

105

1 log t2

1 log10 log52

1 log 22


(a)2

20

s i n x dx1 cos x

(b)2

4 40

sin2 dsin cos

(c)2

0

dx5 4 c o s x

Solution : (a) Let cos x = t then sinxdx= dt

When x = 0, t =1 and x , t 02

. As x varies from 0 to 2

, t varies from 1 to 0.

02 01 12 20 1

sinx 1dx dt tan t1 cos x 1 t

1 1tan 0 tan 1

MATHEMATICS 423

Notes

MODULE - VCalculus

Definite Integrals

04

4

(b)2 2

4 4 22 2 2 20 0

sin2 s i n 2I d dsin cos sin cos 2sin cos

2

2 20

sin 2 d1 2sin cos

2

2 20

sin2 d1 2sin 1 sin

Let 2sin tThen 2sin cos d dt i.e. s in2 d dt

When 0,t 0and , t 12

. As varies from 0 to 2

, the new variable t varies from

0 to 1.1

0

1I dt1 2t 1 t

120

1dt

2t 2t 11

20

1 1I dt1 12 t t4 4

1

2 20

1 1I dt2 1 1t

2 2

1

1

0

1t1 1 2. tan1 122 2

1 1tan 1 tan 1

MATHEMATICS

Notes

MODULE - VCalculus

424

Definite Integrals

4 4 2

(c) We know that 2

2

x1 tan2cosx x1 tan2

2

0

1 dx5 4 c o s x

2

20

2 x2

1 dxx4 1 tan25

1 tan

2 x222 x

0 2

secdx

9 tan(1)

Letxtan t2

Then 2 xsec dx 2dt2

when x 0 , t 0 , when x2

, t 1

12

20 0

1 1dx 2 dt5 4 c o s x 9 t

[From (1)]

11

0

2 ttan3 3

12 1tan

3 3

27.3 SOME PROPERTIES OF DEFINITE INTEGRALSThe definite integral of f (x) between the limits a and b has already been defined as

b

af x dx F b F a ,Where

d F x f x ,dx

where a and b are the lower and upper limits of integration respectively. Now we state belowsome important and useful properties of such definite integrals.

(i)b b

a af x dx f t dt (ii)

b a

a bf x dx f x dx

(iii)b c b

a a cf x dx f x dx f x dx, where a<c<b.

MATHEMATICS 425

Notes

MODULE - VCalculus

Definite Integrals

(iv)b b

a af x dx f a b x dx

(v)2a a a

0 0 0f x dx f x dx f 2a x dx

(vi)a a

0 0f x dx f a x dx

(vii)

2aa

00

0, if f 2a x f x

f x dx2 f x dx, if f 2a x f x

(viii)

aa

a0

0, if f x isanoddfunctionof x

f x dx2 f x dx, if f x isanevenfunctionof x

Many of the definite integrals may be evaluated easily with the help of the above stated proper-ties, which could have been very difficult otherwise.

The use of these properties in evaluating definite integrals will be illustrated in the followingexamples.


(a)2

0log | t a n x | d x 0 (b)

0

x dx1 s i n x

Solution : (a) Let2

0I log | t a n x | d x ....(i)

Using the property a a

0 0f x dx f a x dx,weget

2

0I log tan x dx

2

2

0log cotx dx

MATHEMATICS

Notes

MODULE - VCalculus

426

Definite Integrals

21

0log tan x dx

2

0logtanxdx

I [Using (i)]2I 0

i.e. I = 0 or2

0log | t a n x | d x 0

(b)0

x dx1 s i n x

Let0

xI dx1 s i n x (i)

0xI dx

1 sin x

a a

0 0f x dx f a x dx∵

0

x dx1 s i n x (ii)

Adding (i) and (ii)

0 0

x x 12I dx dx1 sinx 1 sin x

or 20

1 s i n x2I dx1 sin x

2

0sec x tanxsecx dx

0tanx sec x

tan sec tan0 sec0

0 1 0 1

2I

MATHEMATICS 427

Notes

MODULE - VCalculus

Definite Integrals

Example 27.9 Evaluate

(a)2

0

s i n x dxsinx cosx

(b)2

0

sinx cos x dx1 s inxcosx

Solution : (a) Let2

0

sin xI dxsinx cosx

(i)

Also2

0

sin x2I dx

sin x cos x2 2

(Using the property a a

0 0f x dx f a x dx ).

= 2

0

cosx dxcosx sin x

(ii)

Adding (i) and (ii), we get

2

0

sinx cosx2I dxsinx cosx

2

01.dx

20x

2

I4

i.e.2

0

s i n x dx4sinx cosx

(b) Let 2

0

sinx c o s xI dx1 s inxcosx

(i)

MATHEMATICS

Notes

MODULE - VCalculus

428

Definite Integrals

Then

2

0

sin x cos x2 2I dx

1 sin x cos x2 2

a a

0 0f x dx f a x dx∵

2

0

cosx s i n x dx1 cosxs inx

(ii)


2 2

0 0

sinx cosx cosx s i n x2I dx1 sinxcosx 1 s inxcosx

2

0

sinx cosx cosx sin x dx1 s inxcosx

= 0

I = 0

Example 27.10 Evaluate (a) 2a x

2a

xe dx1 x (b)

3

3x 1 dx

Solution : (a) Here2x

2xef x

1 x2x

2xef x

1 xf x

f x is an odd function of x.2a x

2a

xe dx 01 x

(b)3

3x 1 dx

x 1,if x 1x 1

x 1,if x 1

3 1 3

3 3 1x 1 dx x 1 dx x 1 dx, using property (iii)

MATHEMATICS 429

Notes

MODULE - VCalculus

Definite Integrals

1 3

3 1x 1 dx x 1 dx

1 32 2

3 1

x xx x2 2

1 9 9 11 3 3 1 102 2 2 2


0log sinx dx

Solution : Let2

0

I log sinx dx (i)

Also2

0I log sin x dx,

2[Using property (iv)]

2

0log cosx dx (ii)


2

02I log sin x log cosx dx

2

0log sinxcosx dx

2

0

sin2xlog dx2

2 2

0 0log sin2x dx log 2 dx

2

0log sin2x dx log 2

2(iii)

MATHEMATICS

Notes

MODULE - VCalculus

430

Definite Integrals

Again, let2

10

I log sin2x dx

Put 2x = t1dx dt2

When x = 0, t = 0 and x , t2

10

1I log sint dt2

2

0

1 .2 log sint dt2

, [using property (vi)]

2

0

1 .2 log sinx dt2

[using property (i)]

1I I, [from (i)] .....(iv)Putting this value in (iii), we get

2I I log22

I log 22

Hence,2

0log sin x dx log2

2

CHECK YOUR PROGRESS 27.2Evaluate the following integrals :

1.1 2x

0xe dx 2.

2

0

dx5 4 s i n x

3.1

20

2x 3 dx5x 1

4.5

5x 2 dx 5.

2

0x 2 xdx 6.

2

0

s i n x dxcosx s i n x

7.2

0log cosxdx 8.

4a 3 x

2a

x e dx1 x 9.

2

0sin2xlogtanxdx

10.2

0

cosx dx1 sin x cosx

MATHEMATICS 431

Notes

MODULE - VCalculus

Definite Integrals

Fig.27.6

27.4 APPLICATIONS OF INTEGRATION

Suppose that f and g are two continuous functions on an interval [a, b] such that f x g x

for x [a, b] that is, the curve y = f (x) does not cross under the curve y = g (x) over [a, b ].Now the question is how to find the area of the region bounded above by y = f (x), below by y= g (x), and on the sides by x = a and x = b.

Again what happens when the upper curve y = f (x) intersects the lower curve y = g (x) at eitherthe left hand boundary x = a , the right hand boundary x = b or both?

27.4.1 Area Bounded by the Curve, x-axis and the Ordinates

Let AB be the curve y = f (x) and CA, DB the two ordinates at x = a and x = b respectively.Suppose y = f (x) is an increasing function of x inthe interval a x b .

Let P (x, y) be any point on the curve andQ x x, y y a neighbouring point on it.Draw their ordinates PM and QN.

Here we observe that as x changes the area(ACMP) also changes. Let

A=Area (ACMP)

Then the area (ACNQ) A A .

The area (PMNQ)=Area (ACNQ) Area (ACMP)

A A A A.

Complete the rectangle PRQS. Then the area (PMNQ) lies between the areas of rectanglesPMNR and SMNQ, that is

A lies betweeny x and y y x

Ax

lies between y and y y

In the limiting case when Q P, x 0and y 0.

x 0

Alimx

lies between y and y 0lim y y

dA ydx

Integrating both sides with respect to x, from x = a to x = b, we have

MATHEMATICS

Notes

MODULE - VCalculus

432

Definite Integrals

b bba

a a

dAydx dx Adx

= (Area when x = b) (Area when x = a) = Area (ACDB) 0 = Area (ACDB).

Hence Area (ACDB) b

af x dx

The area bounded by the curve y = f (x), the x-axis and the ordinates x = a, x = b is

b

af x dx or

b

aydx

where y = f (x) is a continuous single valued function and y does not change sign in the intervala x b .

Example 27.12 Find the area bounded by the curve y = x, x-axis and the lines x =0, x = 2.

Solution : The given curve is y = x

Required area bounded by the curve, x-axis andthe ordinates x 0, x 2 (as shown in Fig. 27.7)

is2

0xdx

22

0

x2

2 0 2 square units

Example 27.13 Find the area bounded by the

curve xy e , x-axis and the ordinates x = 0 and x = a > 0.

Solution : The given curve is xy e .

Required area bounded by the curve, x-axis and the ordinates x = 0, x = a is

ax

0e dx

ax0

e

ae 1 square units

Fig. 27.7

MATHEMATICS 433

Notes

MODULE - VCalculus

Definite Integrals

Example 27.14 Find the area bounded by the curve x

y ccosc , x-axis and the ordinates

x = 0, x = a, 2a c .

Solution : The given curve is xy ccos

c

Required area a

0ydx

a

0

xccos dxc

a2

0

xc sinc

2 ac sin s in0

c

2 ac sin

c square units

Example 27.15 Find the area enclosed by the circle 2 2 2x y a , and x-axis in the firstquadrant.

Solution : The given curve is 2 2 2x y a , whichis a circle whose centre and radius are (0, 0) and arespectively. Therefore, we have to find the areaenclosed by the circle 2 2 2x y a , the x-axisand the ordinates x = 0 and x = a.

Required area a

0

ydx

a2 2

0a x dx ,

(∵ y is positive in the first quadrant)a2

2 2 1

0

x a xa x sin2 2 a

2 21 1a a0 sin 1 0 sin 0

2 22a .

2 21 1sin 1 ,sin 0 0

2∵

2a4

square units

Fig. 27.8

MATHEMATICS

Notes

MODULE - VCalculus

434

Definite Integrals

Example 27.16 Find the area bounded by the x-axis, ordinates and the following curves :

(i) 2xy c , x a ,x b, a b 0

(ii) ey log x,x a ,x b, b a 1

Solution : (i) Here we have to find the area bounded by the x-axis, the ordinates x = a, x = band the curve

2xy c or2cy

x

Area a

bydx (∵a > b given)

a 2

b

c dxx

a2bc log x

2c loga logb

2 ac log

b(ii) Here ey log x

Area b

ea

log xdx , b a 1∵

bb

e aa

1x log x x dxx

b

e ea

blog b a log a dx

be e ablog b a log a x

e eblog b a log a b a

e eb log b 1 a log a 1

e eb a

blog a loge e elog e 1∵


1. Find the area bounded by the curve 2y x , x-axis and the lines x = 0, x =2.2. Find the area bounded by the curve y = 3 x, x-axis and the lines x = 0 and x = 3.

MATHEMATICS 435

Notes

MODULE - VCalculus

Definite Integrals

3. Find the area bounded by the curve 2xy e , x-axis and the ordinates x = 0, x = a, a > 0.

4. Find the area bounded by the x-axis, the curve x

y c sinc and the ordinates x = 0

and x = a, 2a c .

27.4.2. Area Bounded by the Curve x = f (y) between y-axis and the Lines y = c, y = d

Let AB be the curve x = f (y) and let CA, DBbe the abscissae at y = c, y = d respectively.

Let P (x, y) be any point on the curve and letQ x x, y y be a neighbouring pointon it. Draw PM and QN perpendiculars ony-axis from P and Q respectively. As y changes,the area (ACMP) also changes and hence clearlya function of y. Let A denote the area (ACMP),then the area (ACNQ) will be A A .

The area (PMNQ) = Area (ACNQ) Area (ACMP) A A A A .

Complete the rectangle PRQS. Then the area (PMNQ) lies between the area (PMNS) and thearea (RMNQ), that is,

A lies between x y and x x y

Ay lies between x and x + x

In the limiting position when Q P, x 0 and y 0 .

y 0

Alimy lies between x and x 0

lim x x

dA xdy

Integrating both sides with respect to y, between the limits c to d, we getd d

c c

dAxdy dydy

= dcA

= (Area when y = d) (Area when y = c) = Area (ACDB) 0 = Area (ACDB)

Hence area (ACDB) d d

c cxdy f y dy

Fig. 27.9

MATHEMATICS

Notes

MODULE - VCalculus

436

Definite IntegralsThe area bounded by the curve x = f ( y ), the y-axis and the lines y = c and y = d is

d

cxdy or

d

cf y dy

where x = f ( y ) is a continuous single valued function and x does not change sign in the intervalc y d.

Example 27.17 Find the area bounded by the curve x = y, y-axis and the lines y = 0, y = 3.

Solution : The given curve is x = y.

Required area bounded by the curve, y-axis and the lines y =0, y = 3 is3

0x dy

3

0y dy

32

0

y2

9 02

92

square units

Example 27.18 Find the area bounded by the curve 2x y , y = axis and the lines y = 0, y= 2.

Solution : The equation of the curve is 2x y

Required area bounded by the curve, y-axis and the lines y =0, y = 2

22

0y dy

23

0

y3

8 03

83

square units

Example 27.19 Find the area enclosed by the circle 2 2 2x y a and y-axis in the firstquadrant.

Fig. 27.10

MATHEMATICS 437

Notes

MODULE - VCalculus

Definite Integrals

Solution : The given curve is 2 2 2x y a , which is a circle whose centre is (0, 0) and radius

a. Therefore, we have to find the area enclosed by the circle 2 2 2x y a , the y-axis and theabscissae y = 0, y = a.

Required area a

0x dy

a2 2

0a y dy

(because x is positive in first quadrant)

a22 2 1

0

y a ya y sin2 2 a

2 21 1a a0 sin 1 0 sin 0

2 2

2a4

square units 1 1sin 0 0,sin 12

∵

Note : The area is same as in Example 27.14, the reason is the given curve is symmetricalabout both the axes. In such problems if we have been asked to find the area of the curve,without any restriction we can do by either method.

Example 27.20 Find the whole area bounded by the circle 2 2 2x y a .

Solution : The equation of the curve is 2 2 2x y a .

The circle is symmetrical about both the axes, so the wholearea of the circle is four times the area os the circle in the firstquadrant, that is,Area of circle = 4 × area of OAB

2a44

(From Example 27.15 and 27.19) 2a

square units

Example 27.21 Find the whole area of the ellipse2 2

2 2x y

1a b

Solution : The equation of the ellipse is2 2

2 2x y

1a b

Fig. 27.11

Fig. 27.12

MATHEMATICS

Notes

MODULE - VCalculus

438

Definite IntegralsThe ellipse is symmetrical about both the axes and so the whole area of the ellipse is four times the area in the first quadrant, that is,Whole area of the ellipse = 4 × area (OAB)

In the first quadrant,2 2

2 2y x

1b a

or 2 2by a xa

Now for the area (OAB), x varies from0 to a

Area (OAB) a

0ydx

a2 2

0

b a x dxa

a22 2 1

0

b x a xa x sina 2 2 a

2 21 1b a a

0 sin 1 0 sin 0a 2 2

ab4

Hence the whole area of the ellipse

ab44

ab. square units

27.4.3 Area between two CurvesSuppose that f (x) and g (x) are two continuous and non-negative functions on an interval [a, b]such that f x g x for all x [a, b]that is, the curve y = f (x) does not cross underthe curve y = g (x) for x [a,b] . We want tofind the area bounded above by y = f (x),below by y = g (x), and on the sides by x = aand x = b.

Let A= [Area under y = f (x)] [Area undery = g (x)] .....(1)

Now using the definition for the area boundedby the curve y f x , x-axis and the ordinates x = a and x = b, we have

Area under

Fig.27.13

Fig. 27.14

MATHEMATICS 439

Notes

MODULE - VCalculus

Definite Integrals

b

ay f x f x dx .....(2)

Similarly,Area under b

ay g x g x dx .....(3)

Using equations (2) and (3) in (1), we getb b

a aA f x dx g x dx

b

af x g x dx .....(4)

What happens when the function g has negative values also? This formula can be extended bytranslating the curves f (x) and g (x) upwards until both are above the x-axis. To do this let-m bethe minimum value of g (x) on [a, b] (see Fig. 27.15).

Since g x m g x m 0

Fig. 27.15 Fig. 27.16

Now, the functions g x m and f x m are non-negative on [a, b] (see Fig. 27.16). Itis intuitively clear that the area of a region is unchanged by translation, so the area A between fand g is the same as the area betweeng x m and f x m . Thus,

A = [ area under y f x m ] [area under y g x m ] .....(5)

Now using the definitions for the area bounded by the curve y = f (x), x-axis and the ordinates x= a and x = b, we have

Area under b

ay f x m f x m dx .....(6)

MATHEMATICS

Notes

MODULE - VCalculus

440

Definite Integrals

and Area under b

ay g x m g x m dx (7)

The equations (6), (7) and (5) giveb b

a aA f x m dx g x m dx

b

af x g x dx

which is same as (4) Thus,If f (x) and g (x) are continuous functions on the interval [a, b], andf (x) g (x), x [a, b] , then the area of the region bounded above by y = f (x), belowby y = g (x), on the left by x = a and on the right by x = b is

b

af x g x dx

Example 27.22 Find the area of the region bounded above by y = x + 6, bounded below by2y x , and bounded on the sides by the lines x = 0 and x = 2.

Solution : y x 6 is the equation of the straight line and 2y x is the equation of theparabola which is symmetric about the y-axis and origin the vertex. Also the region is boundedby the lines x 0 and x 2 .

Fig. 27.17

Thus,2 2

2

0 0A x 6 dx x dx

22 3

0

x x6x2 3

34 03

MATHEMATICS 441

Notes

MODULE - VCalculus

Definite Integrals

343

square units

If the curves intersect then the sides of the region where the upper and lower curves intersectreduces to a point, rather than a vertical line segment.

Example 27.23 Find the area of the region enclosed between the curves 2y x and

y x 6 .

Solution : We know that 2y x is the equation of the parabola which is symmetric about they-axis and vertex is origin and y = x + 6 is the equation of the straight line which makes an angle45° with the x-axis and having the intercepts of 6 and 6 with the x and y axes respectively..(See Fig. 27.18).

Fig. 27.18

A sketch of the region shows that the lower boundary is 2y x and the upper boundary is y =x +6. These two curves intersect at two points, say A and B. Solving these two equations we get

2x x 6 2x x 6 0x 3 x 2 0 x 3, 2

When x = 3, y = 9 and when x = 2, y = 4

The required area3 22

x 6 x dx

32 3

2

x x6x2 3

27 222 3

1256

square units

Example 27.24 Find the area of the region enclosed between the curves 2y x and y = x.

Solution : We know that 2y x is the equation of the parabola which is symmetric about the

MATHEMATICS

Notes

MODULE - VCalculus

442

Definite Integralsy-axis and vertex is origin. y = x is the equation of the straight line passing through the origin andmaking an angle of 45° with the x-axis (see Fig. 27.19).

A sketch of the region shows that the lower boundary is 2y x and the upper boundary is theline y = x. These two curves intersect at two points O and A. Solving these two equations, weget

2x x

x x 1 0

x 0,1

Here 2f x x , g x x , a 0andb 1

Therefore, the required area1 20

x x dx

12 3

0

x x2 3

1 12 3

16

square units

Example 27.25 Find the area bounded by the curves 2y 4x and y = x.

Solution : We know that 2y 4x the equation of the parabola which is symmetric about thex-axis and origin is the vertex. y = x is the equation of the straight line passing through origin andmaking an angle of 45° with the x-axis (see Fig. 27.20).

A sketch of the region shows that the lower boundary is y = x and the upper boundary is 2y 4x.These two curves intersect at two points O and A. Solving these two equations, we get

2y y 04

y y 4 0y 0 , 4

When y = 0, x = 0 and when y = 4, x = 4.

Here12

f x 4 x , g x x , a 0 , b 4Therefore, the required area is

142

02x x dx

43 22

0

4 xx3 2

Fig. 27.19

Fig. 27.20

MATHEMATICS 443

Notes

MODULE - VCalculus

Definite Integrals

32 83

83

square units

Example 27.26 Find the area common to two parabolas 2x 4ay and 2y 4ax .

Solution : We know that 2y 4ax and 2x 4ay are the equations of the parabolas, whichare symmetric about the x-axis and y-axis respectively.

Also both the parabolas have their vertices at the origin (see Fig. 27.19).

A sketch of the region shows that the lower boundary is 2x 4ay and the upper boundary is2y 4ax. These two curves intersect at two points O and A. Solving these two equations, we

have4

2x

4ax16a

3 3x x 64a 0

x 0,4a

Hence the two parabolas intersect at point

(0, 0) and (4a, 4a).

Here2xf x 4ax,g x , a 0andb 4a

4a

Therefore, required area

4a 2

0

x4ax dx4a

4a332

0

2.2 ax x3 12a

2 232a 16a3 3

216 a3

square units

Fig. 27.21

MATHEMATICS

Notes

MODULE - VCalculus

444

Definite Integrals

CHECK YOUR PROGRESS 27.41. Find the area of the circle 2 2x y 9

2. Find the area of the ellipse 2 2x y 1

4 9

3. Find the area of the ellipse 2 2x y 1

25 16

4. Find the area bounded by the curves 2

2 xy 4axandy4a

5. Find the area bounded by the curves 2 2y 4xandx 4y.

6. Find the area enclosed by the curves 2y x and y x 2

LET US SUM UP

If f is continuous in [a, b] and F is an anti derivative of f in [a, b], thenb

af x dx F b F a

If f and g are continuous in [a, b] and c is a constant, then

(i)b b

a ac f x dx c f x dx

(ii)b b b


(iii)b b b


The area bounded by the curve y = f (x), the x-axis and the ordinates

x a , x b isb

af x dx or

b

aydx

where y f x is a continuous single valued function and y does not change sign inthe interval a x b

LET US SUM UP

MATHEMATICS 445

Notes

MODULE - VCalculus

Definite Integrals

If f (x) and g (x) are continuous functions on the interval [a, b] and f x g x , for all

x a,b , then the area of the region bounded above by y = f (x), below by y = g (x), onthe left by x = a and on the right by x = b is

b

af x g x dx


TERMINAL EXERCISE

Evaluate the following integrals (1 to 5) as the limit of sum.

1.b

axdx 2.

b2

ax dx 3.

b

asinxdx

4. b

acosxdx 5.

22

0x 1 dx

Evaluate the following integrals (6 to 25)

6.2

2 2

0a x dx 7.

2

0sin2xdx 8.

2

4

cotxdx

9.2

2

0cos xdx 10.

11

0sin xdx 11.

1

20

1 dx1 x

12.4

23

1 dxx 4 13.

0

1 d5 3cos 14.

43

02tan xdx

15.2

3

0sin xdx 16.

2

0x x 2dx 17.

25

0sin cos d


MATHEMATICS

Notes

MODULE - VCalculus

446

Definite Integrals

18.0

xlogsinxdx 19.0

log 1 cosx dx 20. 20

x sin x dx1 cos x

21.2 2

0

sin x dxsinx cosx

22.4

0log 1 tan x dx 23.

85

0sin 2xcos2xdx

24.2 32

0x x 1 dx

MATHEMATICS 447

Notes

MODULE - VCalculus

Definite Integrals

ANSWERS


1.352

2.1ee

3. (a)2 1

2(b) 2 (c)

4(d)

643


1.e 1

22. 12 1tan

3 33. 11 3

log6 tan 55 5

4. 29 5.24 2

156.

47. log2

2

8. 0 9. 0 10.1

log22 2


1.83

sq. units 2.272

sq. units 3.2ae 1

2sq. units

4. 2 ac 1 cos

c

CHECK YOUR PROGRESS 27.41. 9 sq. units 2. 6 sq. units 3. 20 sq. units

4. 216 a3

sq. units 5.163

sq. units 6.92

sq. units

TERMINAL EXERCISE

1.2 2b a

22.

3 3b a3

3. cosa cosb

4. sinb s ina 5.143

6.2a

4

MATHEMATICS

Notes

MODULE - VCalculus

448

Definite Integrals

7. 1 8.1 log22

9.4

10. 12

11.2

12.1 5log4 3

13.4

14. 1 log 2 15.23

16.16 2 215

17.64231

18.2

log 22

19. log 2 20.2

421.

1log 1 2

2

22. log28

23.1

9624. 78

448/1

27.5 Reduction formulae

In this section, we derive some reduction formulae for the evaluation of definite integrals of

sinnx, cosnx, sinmx, cosnx between 0 and 2π

for positive integers m, n.

The following theorem gives a useful formula to evaluate the definite integral of sinnx between 0

and2π

, when n is an integer n > 2.

27.5.1 Theorem

Let n be an integer greater than or equal to 2. Then

/ 2

0

1 3 1...... , If is even

2 2 2sin1 3 2

...... , If is odd2 3

n

n nn

n nx dxn n

nn n

π− − π

−= − − −

∫

Note: 1./ 2 / 2

0 0

sin cosn nx dx x dxπ π

=∫ ∫

Example 1. Find

(1)

/ 24

0

sin x dxπ

∫

(ii)/ 2

7

0

sin x dxπ

∫

(iii)/ 2

8

0

cos x dxπ

∫

448/2

Sol: (i) / 2

4

0

4 1 4 3sin . .

4 4 2 2x dx

π − − π=−∫

3 1 3. .

4 2 2 16

π= = π

(ii) / 2

7

0

7 1 7 3 7 5sin . .

7 7 2 7 4x dx

π − − −=− −∫

6 4 2 16. .

7 5 3 35= = .

(iii) / 2

8

0

8 1 8 3 8 5 8 7cos . . . .

8 8 2 8 4 8 6 2x dx

π − − − − π=− − −∫

7 5 3 1. . . .

8 6 4 2 2

π=

35.

256= π

27.5.2 Theorem : Let m and n be positive integers. Then

1 3 2 1. ... . . If 1 is odd

2 3 1

n nn

m n m n m m

− − ≠+ + − + +

/ 2

0

sin cosm nx x dxπ

=∫ 1 3 1 1 1. ... . .... .

2 2 2 2

n n m

m n m n m m

− − − π+ + − +

If n is even and m is even

1 3 1 1 2. ... . ....

2 2 3

n n m

m n m n m m

− − −+ + − +

If n is even and 1 m≠ is odd.

Example 2

Evaluate the following definite integrals

(i)/ 2

4 5

0

sin cosx xdxπ

∫

(ii)/ 2

5 4

0

sin cosx xdxπ

∫

448/3

(iii)/ 2

6 4

0

sin cosx xdxπ

∫

Sol :

(i)/ 2

4 5

0

5 1 5 3 1sin cos . .

4 5 4 5 2 4 1x xdx

π − −=+ + − +∫

4 2 1 8. .

9 7 5 315= =

(ii)/ 2

5 4

0

4 1 4 3 5 1 5 3sin cos . . .

5 4 5 4 2 5 5 2x x dx

π − − − −=+ + − −∫

3 1 4 2 8. . .

9 7 5 3 315= =

(iii)/ 2

6 4

0

4 1 4 3 6 1 6 3 6 5sin cos . . . . .

6 4 6 4 2 6 6 2 6 4 2x x dx

π − − − − − π=+ + − − −∫

3 1 5 3 1. . . . .

10 8 6 4 2 2

π=

3.

512= π

Example 3.

Find2

4 6

0

sin cosx x dxπ

∫Sol: Let f(x) = sin4x cos6x

Since f(2π −x) = f(π −x) = f(x), it follows from previous theorem

24 6

0 0

( ) 2 sin cosf x dx x xdxπ π

=∫ ∫

/ 24 6

0

4 sin cosx x dxπ

= ∫

6 1 6 3 6 5 3 14. . . . . .

4 6 4 6 2 4 6 4 4 2 2

− − − π=+ + − + −

3.

128= π

448/4

Example 4

Find / 2

2 4

/ 2

sin cosx x dxπ

−π∫

Sol: Let f(x) = sin2x cos4x since f is even we have

/ 2 / 2

/ 2 0

( ) 2 ( )f x dx f x dxπ π

−π

=∫ ∫

Hence

/ 2 / 22 4 2 4

/ 2 0

sin cos 2 sin cosx x dx x x dxπ π

−π

=∫ ∫

4 1 4 3 12. . . .

2 4 2 4 2 2 2

− − π=+ + −

3 12. . . .

6 4 4 16

π π= =

Check Your Progress

I. Find the values of the following integrals.

1./ 2

10

0

sin x dxπ

∫

2./ 2

4 4

0

sin cosx x dxπ

∫

3.

/ 27 2

0

cos sinx x dxπ

∫

4./4

3

0

sin dπ

θ θ∫

Let Us Sum Up

1. If n is an integer n > 2, then

/ 2

0

1 3 1...... . , If is even

2 2 2sin1 3 2

...... . If is odd2 3

n

n nn

n nx dxn n

nn n

π− − π

−= − −

−

∫

448/5

2./ 2 / 2

3

0 0

sin cos ,nx dx x dxπ π

=∫ ∫ n is a positive integer.

3. If m and n are positive intgers then

/ 2

0

sin cosm nx x dxπ

=∫ 1 3 2 1. ... . .

2 3 1

n n

m n m n m m

− −+ + − + + If 1 n≠ is odd

1 3 1 1 1. ... . ....

2 2 2 2

n n m

m n m n m m

− − − π+ + − +

If n is even and m is even

1 3 1 1 2. ... . ....

2 2 3

n n m

m n m n m m

− − −+ + − +

If n is even and 1 m≠ is odd.

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Answers

Check Your Progress

(1)63

512π

(2)3

256π

(3)16

315

(4)2 5

3 6 2−

448/6

Numerical Integration

Introduction

Integral calculus arose from the ara problem. Before Newton, and Leibnitz developed a system-

atic method to evaluate a difinite integral bassed on the Fundamental theorem of calculus, problems of

finding areas, volumes and lengths of curves were so difficult that only a genius could attempt to solve

tham. But now, one can solve such challenging problems with greater ease. In this chapter we study

the area bounded by the graph of a non-negative continuous function y = f(x) defined on a closed

interval [a, b] and the x-axis as an application of the definite integral.

Objectives

From this chapter we can get to know the following points.

� The area bounded between the two curves

� lengths, volumes and areas of the curve

� When the integration becomes difficult we can find the approximate value with simpson or

Trapezoidel rule.

27.6 Determination of plane areas involving second degree curves

In chapter 9, it was observed that if y = f (x) is non-negative continuous function defined on

[a, b] , then the area under the graph of f betwen the ordinates x = a, x = b and the X-axis is given

by the value of the definite integral ( )b

a

f x dx∫ . In this section we will give different possible ways of

calculating such areas depending on the nature of the integrand.

448/7

Y

O X

y = f(x)

a b

A

Y

O X

y = f(x)

a b

A

27.6.1 Area under curves

(i) If f : [a, b] [0, )→ ∞ is continuous, then the area A of the region bounded by the curve

y = f(x), the X-axis and the lines x = a and x = b is given by A ( )b

a

f x dx=∫ which is shown in

fig27.2 graphically.

Fig. 27.2

(ii) Let f : [a, b] ( , 0]→ −∞ be continuous. Then the graphs of y = f(x) and y = −f(x) in [a, b]

are symmetric about the X-axis. So, the area bounded by the graph of y = f(x), the X-axis and

the lines x = a, x = b is same as the area bounded by the graph of y = −f(x), the X-axis and the

lines x = a, x = b which is, hence, given by A ( )b

a

f x dx= −∫ . This is shown graphically in the

fig. 27.3.

From (i) and (ii) we observe that A ( )b

a

f x dx= ∫

Fig. 27.3

448/8

Y

O X

y = f(x)

a bc

A

(iii) Let f : [a , b] R→ be continuous and f(x) > 0 for all x∈ [a , c] and f(x) < 0 for all x∈ [c, b]

where a < c < b. Then the area of the region bounded by the curve y = f(x) , X-axis, and the

lines. x = a, x = b is given byb

A ( ) ( )c b

a c

f x dx f x dx= + −∫ ∫

( ) ( )c b

a c

f x dx f x dx= −∫ ∫ ...(1)

Which is shown in fig 27.4 graphically.

Fig. 27.4

Note: (1) can be written as A ( ) ( )c b

a c

f x dx f x dx= +∫ ∫

(iv) Let : [ , ] R , : [ , ] Rf a b g a b→ → be continuous and ( ) ( )f x g x≤ for all x∈ [a, b]. Then

the area of the region bounded by the curves y = f(x), y = g(x)]and the lines x = a, x = b is given

by

A ( ) ( )b b

a a

g x dx f x dx= −∫ ∫

In case ( ) ( )g x f x≤ for all x∈ [a, b] then the area A is given by

A ( ) ( )b b

a a

f x dx g x dx= −∫ ∫

A ( ) ( )b b

a a

f x dx g x dx= −∫ ∫

448/9

Fig. 27.5


Find the area of the region enclosed by the given curves.

I 1. y = cos x,2

1x

y = −π

2. 3 3, 0, 1, 2y x y x x= + = = − =

3. 24 , 0x y x= − =

II (1) 2 4 , 2 , 0x y x y= = =

(2) 2 3 , 3y x x= =

(3) 2 , 2y x y x= =

27.7 Trapezoidal rule and Simpson's rule - Simple applications

As already explained in the introduction, sometimes it is extremely difficult or even impossible to

evaluate a definite integral ( )b

a

f x dx∫ even if f is continuous on [a, b]. In such cases we take a set

of numerical values of the integrand f in the interval [a,b] and evaluate the definite integral ( )b

a

f x dx∫approximately. This process of finding the approximate value of a definite integral is called Numerical

Integration. If the integrand is a function of single variable, then this process is known as quadrature.

Y

O X

y = g(x)

a b

y = f(x)

A

448/10

O Xa bx1 x2 xr−1 xr xn−1

Y

There are several methods for approximating a definite integral. We discuss only two methods

namely Trapezoidal rule and simpson's rule.

27.7.1 The Trapezoidal rule

Let f : [a, b] [0, )→ ∞ be continuous and 0 1 2P { , , ... }na x x x x b= = = be a partition of

[a, b] into n equal subintervals, each of length b ah

n

−= Let y0, y1, y2 ... yn be the values of f at

0 1 2, , ... nx x x x x= respectively (see fig. 27.6).

Fig. 27.6

( ) ( ) ( )0 1 1 2 1( ) ...2 2 2

b

n n

a

h h hf x dx y y y y y y−≅ + + + + + +∫

( ) ( )0 1 2 12 ...2 n nh

y y y y y− = + + + + +

i.e., ( )2

b

a

hf x dx ≅∫ [(Sum of the first and last ordinates) + 2(Sum of the remaining ordinates)]

27.7.2 Working rule for Trapezoidal rule

Step1 : Note down a, b and n, the number of divisions

Step 2 : Calculate b ah

n

−=

448/11

Step 3 : Compute the values of f at x = x0( = a), x = x1, x = x2 ..., x = xn (= b) and denote them

as y0 , y1 , y2 ... yn. respectively.

Step 4 : Write down the Trapezoidal rule for the division specified in the question and substitute the

values obtained in Step 3 in the rule (1) and sumplify.

Example 1

Find the approximate vale of 9

2

1

x dx∫ using trapezoidal rule with 4 equal intervals.

Sol : n = 4, a = 1, b = 9

9 12

4h

−= =

y = x2

x 1 3 5 7 9

y 1 9 25 49 81

Required Area 2[(1 81) 2 (9 25 49)]

2= + + + +

= 82 + 166

= 248

27.7.3 Simpson's rule of evaluation of ( )b

a

f x dx∫ approximately

This is another rule for approximate integration results using parabolas instead of straight line

segments to approximate a curve. As before, we divide [a, b] into n subintervals of equal length

b ah

n

−= , but this time we assume that n is even. Here we approximate the curve y = f(x) > 0

passing through three successive points of subdivision by a parabola as shown in fig 27.7.

Let yr = f(xr) for r = 0, 1, 2, ... n; and Pr be the point on the curve (xr, yr).

448/12

Y

O Xa=x0 x1 x2 xn−2xn−1x3 xn=b

Pn−1

Pn−2

Pn

P0P1 P2

P3 P4

Fig. 27.7

For simplifying the calculations, we first consider the case where x0 = −h, x1 = 0 and x2 = h (see

fig. 27.8). The equation of the parabola passing through P0, P1 and P2 is of the form

y = Ax2 + Bx + C and therefore the area under the parabola from x = −h to x = h

2 2 2

0

(A B C) 2 (A C) (2A 6C)3

h h

h

hx x dx x dx h

−

+ + = + = +∫ ∫ .

Fig. 27.8

But since the parabola passes through P0(−h, y0), P1(0, y1) and P2(h, y2) we have

y0 = A h2 − Bh + C,

y1 = C

Hence y2 = Ah2 + Bh + C

Thus, the area under the parabola passing through P0, P1 and P2 is 0 1 2( 4 )3h

y y y+ + .

Y

O Xh−h

P0(−h, y0) P2(h, y2)

P1(0, y1)

448/13

Similarly, the area under the parabola through P2, P3 and P4 from x = x2 to x = x4 is

2 3 4( 4 )3h

y y y+ + . If we compute the areas under all the parabolas in the above manner and add the

results, we obtain

0 1 2 1 3 4( ) [( 4 ) ( 4 ) .......3

b

a

hf x dx y y y y y y≅ + + + + + +∫

2 1( 4 ) ]n n ny y y− −+ + + ...(2)

i.e.,

0 1 3 5 1 2 4 6 2( ) [( ) 4( ... ) 2( ... ) ]3

b

n n na

hf x dx y y y y y y y y y y− −≅ + + + + + + + + + +∫

= 3

h [ (Sum of the first and last ordinates ) + 4(Sum of the even ordinates)

+ 2(Sum of all other odd ordinates execluding first and last)]

Although we have derived this approximation for the case ( ) 0f x ≥ , it is a reasonable ap-

proximation for any continuous function f and is called Simpson's Rule or Simpson's one-third rule

after the English mathematician Thomas Simpson.

Example 2

Find the approximate value of π from 1

20

1

1dx

x+∫ using sumpson's rule by dividing [0, 1] in to

4 equal parts.

Sol: n = 4 , 1 0 1

4 4

b ah

n

− −= = =

2

1

1y

x=

+

x 01

4

1

2

3

41

y 116

17

4

5

16

25

1

2

448/14

1

0 2 1 3 2 1 2 4 2 220

[( ) 4( ... ) 2( ... )]31

n n ndx h

y y y y y y y yx

− −= + + + + + + + + ++∫

1 1 16 16 41 4 2

12 2 17 25 5

= + + + +

( )11.5 4 0.9412 0.64 1.6

12 = + + +

[ ]13.1 6.3248

12= +

9.4248

12=

= 0.7854

But 1

11 1 12 0

0

Tan Tan 1 Tan 041

dxx

x− − − π = = − = +∫

0.7854 3.14164

π∴ = ⇒ π =


I. 1. Calculate the approximate value of1

2

0

(1 )x dx+∫ by dividing [0 , 1] into 5 equal parts using

Trapezoidal rule.

2. Calculate the approximate value of 3

1

1dx

x∫ , by suing simpson's rule and taking n = 4.


11dx

dxx+∫ by taking n = 4 in the simpson's rule.

448/15

II. 1. Calculate the approximate value of 3

4

3

x dx−∫ using

(i) Trapezoidal rule (ii) Simpson's rule by dividing [−3, 3] fin to 6 equal parts.

2. Use simson's rule to evaluate 7

1

1dx

x∫ approximately by taking n = 6 and hence find the

approximate value of loge7 to three places of decimals.


2

0

1 x dx+∫ , by taking n = 6 in sumpson's rule.

4. Evaluate1

3

0

1 x dx+∫ approximately by dividing [0, 1] into 10 subintervals of the same

length by (i) Trapezoidal rule (ii) Simposon's rule.

Let Us Sum Up

1. Let f (x) > 0 for all x∈ [a, b] and

p = {a = x0, x1, x2, ....xr, ... xn = b] be a partition of [a, b] into n equal subintervals, each of

lengthb a

hn

−= . If f (xr) = yr, for 0 < r < n then

(i) Trapezoidal rule

0 1 3 5 1 2 4 6 2( ) [( ) 4( ... ) 2 ( ... ) ]3

b

n n na

hf x dx y y y y y y y y y y− −≅ + + + + + + + + + + +∫

2. Simposon's one third rule (for even n)

0 1 3 5 1 2 4 6 2( ) [( ) 4( ... ) 2 ( ... ) ]3

b

n n n

a

hf x dx y y y y y y y y y y− −≅ + + + + + + + + + + +∫

3. 0 1 2 1( ) [( ) 2( ... )3

b

n n

a

hf x dx y y y y y−≅ + + + + +∫

Supportive Websites



448/16

Answers


I. 1. 22

π− 2. 51

4

3. 32

3

II. (1) 2

3

(2) 1 2

(3) 4

3


I. (1) 1.34

(2) 1.1

(3) 1.1

(4) 0.695635

I. (1) 115

(2) 1.959

(3) 19.5011

(4) (i) 1.1122

(ii) 1.11132

MATHEMATICS 449

Notes

MODULE - VCalculus

Differential Equations

28

DIFFERENTIAL EQUATIONS

Having studied the concept of differentiation and integration, we are now faced with the questionwhere do they find an application.In fact these are the tools which help us to determine the exact takeoff speed, angle of launch,amount of thrust to be provided and other related technicalities in space launches.Not only this but also in some problems in Physics and Bio-Sciences, we come across relationswhich involve derivatives.

One such relation could be 2ds 4.9 tdt

where s is distance and t is time. Therefore, dsdt

represents velocity (rate of change of distance) at time t.Equations which involve derivatives as their terms are called differential equations. In this lesson,we are going to learn how to find the solutions and applications of such equations.


define a differential equation, its order and degree;determine the order and degree of a differential equation;form differential equation from a given situation;illustrate the terms "general solution" and "particular solution" of a differential equationthrough examples;solve differential equations of the following types :

(i)dy f ( )dx

x (ii)dy f ( )g ( )dx

x y

(iii)dy f ( )dx g

xy (iv)

dy P y Qdx

x x (v)2

2d y

f( )dx

x

find the particular solution of a given differential equation for given conditions.

MATHEMATICS

Notes

MODULE - VCalculus

450


EXPECTED BACKGROUND KNOWLEDGE

Integration of algebraic functions, rational functions and trigonometric functions

28.1 DIFFERENTIAL EQUATIONSAs stated in the introduction, many important problems in Physics, Biology and Social Sciences,when formulated in mathematical terms, lead to equations that involve derivatives. Equations

which involve one or more differential coefficients such as 2

2dy d y

,dx dx

(or differentials) etc. and

independent and dependent variables are called differential equations.

For example,

(i)dy cosxdx

(ii)2

2d y

y 0dx

(iii) xdx ydy 0

(iv)2 32

22

d y dyx 0dxdx

(vi)2dy dyy 1

dx dx

28.2 ORDER AND DEGREE OF A DIFFERENTIAL EQUATIONOrder : It is the order of the highest derivative occurring in the differential equation.Degree : It is the degree of the highest order derivative in the differential equation after theequation is free from negative and fractional powers of the derivatives. For example,

Differential Equation Order Degree

(i)dy s i n xdx

One One

(ii)2

2dy 3y 5xdx One Two

(iii)2 42

22

d s dst 0dtdt

Two Two

(iv)3

3d v 2 dv

0r drdr

Three One

(v)2 54 3

34 3

d y d yx s i n xdx dx

Four Two

MATHEMATICS 451

Notes

MODULE - VCalculus


Example 28.1 Find the order and degree of the differential equation :

322 2

2d y dy1 0

dxdx

Solution : The given differential equation is3

22 2

2d y dy1 0

dxdxor

322 2

2d y dy

1dxdx

The term

32 2dy1

dx has fractional index. Therefore, we first square both sides to remove

fractional index.

Squaring both sides, we have32 22

2d y dy1

dxdx

Hence each of the order of the diferential equation is 2 and the degree of the differential equationis also 2.

Note : Before finding the degree of a differential equation, it should be free from radicals andfractions as far as derivatives are concerned.

28.3 LINEAR AND NON-LINEAR DIFFERENTIAL EQUATIONSA differential equation in which the dependent variable and all of its derivatives occur only in thefirst degree and are not multiplied together is called a linear differential equation. A differentialequation which is not linear is called non-linear differential equation . For example, the differentialequations

2

2d y

y 0dx

and3

2 33

d y dycos x x y 0

dxdx are linear..

The differential equation 2dy y log x

dx x is non-linear as degree of dydx

is two.

Further the differential equation 2

2d y

y 4dx

x is non-linear because the dependent variable

y and its derivative 2

2d ydx

are multiplied together..

MATHEMATICS

Notes

MODULE - VCalculus

452


28.4 FORMATION OF A DIFFERENTIAL EQUATIONConsider the family of all straight lines passing through the origin (see Fig. 28.1).

This family of lines can be represented byy = mx .....(1)

Differentiating both sides, we getdydx

m .....(2)

From (1) and (2), we getdyydx

x .....(3)

Sodyy x anddx

m y x represent the same family..

Clearly equation (3) is a differential equation.

Working Rule : To form the differential equation corresponding to an equation involvingtwo variables, say x and y and some arbitrary constants, say, a, b, c, etc.(i) Differentiate the equation as many times as the number of arbitrary constants in the

equation.(ii) Eliminate the arbitrary constants from these equations.

Remark

If an equation contains n arbitrary constants then we will obtain a differential equation of nth

order.

Example 28.2 Form the differential equation representing the family of curves.2y ax bx. ......(1)

Differentiating both sides, we getdy 2 ax bdx

......(2)

Differentiating again, we get2

2d y

2 adx

......(3)

2

21 d y

a2 dx

......(4)

(The equation (1) contains two arbitrary constants. Therefore, we differentiate this equationtwo times and eliminate 'a' and 'b').On putting the value of 'a' in equation (2), we get

2

2dy d yx bdx dx

Fig. 28.1

MATHEMATICS 453

Notes

MODULE - VCalculus


2

2dy d y

b xdx dx

......(5)

Substituting the values of 'a' and 'b' given in (4) and (5) above in equation (1), we get2 2

22 2

1 d y dy d yy x x x

2 dxdx dx

or2 2 2

22 2

x d y dy d yy x x

2 dxdx dx

or2 2

2dy x d y

y xdx 2 dx

or2 2

2x d y dy

x y 02 dxdx

which is the required differential equation.

Example 28.3 Form the differential equation representing the family of curves

y = a cos (x + b).

Solution : y = a cos (x +b) .....(1)Differentiating both sides, we get

dy a sin(x b)dx

....(2)


2d y

acos x bdx

.....(3)

From (1) and (3), we get2

2d y

ydx

or2

2d y

y 0dx


Example 28.4 Find the differential equation of all circles which pass through the origin andwhose centres are on the x-axis.Solution : As the centre lies on the x-axis, its coordinates will be (a, 0).Since each circle passes through the origin, its radius is a.Then the equation of any circle will be

2 2 2x a y a (1)

To find the corresponding differential equation, we differentiate equation (1) and get

MATHEMATICS

Notes

MODULE - VCalculus

454


dy2 x a 2y 0dx

ordyx a y 0dx

ordya y xdx

Substituting the value of 'a' in equation (1), we get2 2

2dy dyx y x y y xdx dx

or2 2

2 2dy dy dyy y x y 2xydx dx dx

or 2 2 dyy x 2xydx


RemarkIf an equation contains one arbitrary constant then the corresponding differential equation isof the first order and if an equation contains two arbitrary constants then the correspondingdifferential equation is of the second order and so on.

Example 28.5 Assuming that a spherical rain drop evaporates at a rate proportional to itssurface area, form a differential equation involving the rate of change of the radius of the raindrop.Solution : Let r(t) denote the radius (in mm) of the rain drop after t minutes. Since the radius isdecreasing as t increases, the rate of change of r must be negative.If V denotes the volume of the rain drop and S its surface area, we have

34V r3

.....(1)

and 2S 4 r .....(2)It is also given that

dV Sdt

ordV Sdt

k

ordV dr. Sdr dt

k .....(3)

Using (1), (2) and (3) we have

2 2dr4 r . 4 rdt

k

MATHEMATICS 455

Notes

MODULE - VCalculus


ordrdt

k


CHECK YOUR PROGRESS 28.11. Find the order and degree of the differential equation

dy 1y x dydxdx

2. Write the order and degree of each of the following differential equations.

(a)4 2

2ds d s3s 0dt dt

(b)2dy dyy 2x x 1

dx dx

(c) 2 21 x dx 1 y dy 0 (d)2 32

2d s ds3 4 0

dtdt3. State whether the following differential equations are linear or non-linear.

(a) 2 2xy x dx y x y dy 0 (b) dx dy 0

(c)dy cosxdx

(d) 2dy sin y 0dx

4. Form the differential equation corresponding to2 2 2x a y b r by eliminating 'a' and 'b'.

5. (a) Form the differential equation corresponding to2 2 2y m a x

(b) Form the differential equation corresponding to2 2 2y 2ay x a , where a is an arbitrary constant.

(c) Find the differential equation of the family of curves 2x 3xy Ae Be where Aand B are arbitrary constants.

(d) Find the differential equation of all straight lines passing through the point (3,2).(e) Find the differential equation of all the circles which pass through origin and whose

centres lie on y-axis.

28.5 GENERAL AND PARTICULAR SOLUTIONSFinding solution of a differential equation is a reverse process. Here we try to find anequation which gives rise to the given differential equation through the process ofdifferentiations and elimination of constants. The equation so found is called the primitive orthe solution of the differential equation.

MATHEMATICS

Notes

MODULE - VCalculus

456

Differential Equations Remarks

(1) If we differentiate the primitive, we get the differential equation and if we integrate thedifferential equation, we get the primitive.

(2) Solution of a differential equation is one which satisfies the differential equation.

Example 28.6 Show that 1 2y C sin x C cos x, where 1 2C and C are arbitrary

constants, is a solution of the differential equation :2

2d y

y 0dx


1 2y C sinx C cosx .....(1)Differentiating both sides of (1), we get

1 2dy C cosx C sin xdx

.....(2)


1 22d y

C sinx C cosxdx

Substituting the values of 2

2d ydx

and y in the given differential equation, we get

21 2 1 22

d yy C sin x C cosx ( C sinx C cosx)

dx

or2

2d y

y 0dx

In integration, the arbitrary constants play important role. For different values of the constantswe get the different solutions of the differential equation.

A solution which contains as many as arbitrary constants as the order of the differential equationis called the General Solution or complete primitive.

If we give the particular values to the arbitrary constants in the general solution of differentialequation, the resulting solution is called a Particular Solution.

Remark

General Solution contains as many arbitrary constants as is the order of the differential equation.

Example 28.7 Show that ay cxc

(where c is a constant) is a solution of the differential

equation.

MATHEMATICS 457

Notes

MODULE - VCalculus


dy dxy x adx dy

Solution : We have ay cxc

.....(1)

Differentiating (1), we get

dy cdx

dx 1dy c

On substituting the values ofdydx

and dxdy in R.H.S of the differential equation, we have

1 ax c a cx y

c c R.H.S. = L.H.S.

Henceay cxc

is a solution of the given differential equation.

Example 28.8 If 2y 3x C is the general solution of the differential equation

dy 6x 0dx

, then find the particular solution when y = 3, x = 2.

Solution : The general solution of the given differential equation is given as

2y 3x C ....(1)

Now on substituting y = 3, x = 2 in the above equation , we get3 = 12 + C or C 9

By substituting the value of C in the general solution (1), we get

2y 3x 9

which is the required particular solution.

28.6 TECHNIQUES OF SOLVING A DIFFERENTIAL EQUATION

28.6.1 When Variables are Separable

(i) Differential equation of the type dy f(x)dx

Consider the differential equation of the type dy f(x)dx

or dy = f (x) dxOn integrating both sides, we get

dy f x dx

MATHEMATICS

Notes

MODULE - VCalculus

458


y f x dx c

where c is an arbitrary constant. This is the general solution.

Note : It is necessary to write c in the general solution, otherwise it will become a particularsolution.

Example 28.9 Solve

2dyx 2 x 4x 5dx

Solution : The given differential equation is 2dyx 2 x 4x 5dx

or2dy x 4x 5

dx x 2or

2dy x 4x 4 4 5dx x 2

or2x 2dy 9

dx x 2 x 2or

dy 9x 2dx x 2

or9

dy x 2 dxx 2 .....(1)

On integrating both sides of (1), we have

9dy x 2 dx

x 2 or

2xy 2x 9 log | x 2 | c2

,

where c is an arbitrary constant, is the required general solution.

Example 28.10 Solve

3dy 2x xdx

given that y = 1 when x = 0

Solution : The given differential equation is 3dy 2x xdx

or 3dy 2x x dx ...(1)

On integrating both sides of (1), we get

3dy 2x x dx or4 2x xy 2. C

4 2

or4 2x xy C

2 2.....(2)

where C is an arbitrary constant.Since y = 1 when x = 0, therefore, if we substitute these values in (2) we will get

1 0 0 C C = 1

MATHEMATICS 459

Notes

MODULE - VCalculus

Differential EquationsNow, on putting the value of C in (2), we get

4 21y x x 12

or 2 21y x x 1 12


(ii) Differential equations of the typedy f(x) g(y)dx

Consider the differential equation of the typedy f (x) g(y)dx

ordy f (x) dx

g(y) .....(1)

In equation (1), x's and y's have been separated from one another. Therefore, this equation isalso known differential equation with variables separable.To solve such differential equations, we integrate both sides and add an arbitrary constant onone side.

To illustrate this method, let us take few examples.

Example 28.11 Solve2 21 x dy 1 y dx

Solution : The given differential equation2 21 x dy 1 y dx

can be written as 2 2dy dx

1 y 1 x (Here variables have been seperated)


2 2dy dx

1 y 1 x

or 1 1tan y tan x C

where C is an arbitrary constant.This is the required solution.

Example 28.12 Solve

2 2 2 2dyx yx y xy 0dx

Solution : The given differential equation

MATHEMATICS

Notes

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460


2 2 2 2dyx yx y xy 0dx

can be written as 2 2dyx 1 y y 1 x 0dx

or 2 21 y 1 x

dy dxy x

(Variables separated) ....(1)If we integrate both sides of (1), we get

2 21 1 1 1

dy dxy xy x

where C is an arbitrary constant.

or1 1log y log x Cy x

orx 1 1log Cy x y

Which is the required general solution.

Example 28.13 Find the particular solution of

2dy 2xdx 3y 1

when y (0) = 3 (i.e. when x = 0, y = 3).Solution : The given differential equation is

2dy 2xdx 3y 1 or 23y 1 dy 2xdx (Variables separated) ....(1)

If we integrate both sides of (1), we get23y 1 dy 2xdx ,

where C is an arbitrary constant.3 2y y x C ....(2)

It is given that, y 0 3.on substituting y = 3 and x = 0 in (2), we get

27 3 C C = 30

Thus, the required particular solution is3 2y y x 30

MATHEMATICS 461

Notes

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28.6.2 Homogeneous Differential EquationsConsider the following differential equations :

(i) 2 2 dy dyy x xydx dx

(ii) 3 3 2x y dx 3xy dy 0

(iii)3 2

2dy x xydx y x

In equation (i) above, we see that each term except dydx

is of degree 2

[as degree of 2y is 2, degree of 2x is 2 and degree of xy is 1 + 1 =2]

In equation (ii) each term except dydx

is of degree 3.

In equation (iii) each term except dydx

is of degree 3, as it can be rewritten as

2 3 2dyy x x x ydx

Such equations are called homogeneous equations .

Remarks Homogeneous equations do not have constant terms.

For example, differential equation2 3x 3yx dx x x dy 0

is not a homogeneous equation as the degree of the function except dydx

in each term is not the

same. [degree of 2x is 2, that of 3yx is 2, of 3x is 3, and of x is 1]

28.6.3 Solution of Homogeneous Differential Equation :To solve such equations, we proceed in the following manner :(i) write one variable = v. (the other variable).

( i.e. either y = vx or x = vy )(ii) reduce the equation to separable form(iii) solve the equation as we had done earlier.

Example 28.14 Solve

2 2 2x 3xy y dx x dy 0

Solution : The given differential equation is

2 2 2x 3xy y dx x dy 0

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Notes

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462


or2 2

2dy x 3xy ydx x

.....(1)

It is a homogeneous equation of degree two. (Why?)Let y vx . Then

dy dvv xdx dx

From (1), we have

22

2x 3x.vx vxdvv x

dx xor

22

2dv 1 3v v

v x xdx x

or 2dvv x 1 3v vdx

or 2dvx 1 3v v vdx

or 2dvx v 2v 1dx

or 2dv dx

xv 2v 1

or 2dv dx

xv 1 .....(2)

Further on integrating both sides of (2), we get

1 C log xv 1

, where C is an arbitrary constant.

On substituting the value of v, we getx log x C

y x which is the required solution.

28.6.4 Differential EquationConsider the equation

dy Py Qdx

.....(1)

where P and Q are functions of x. This is linear equation of order one.

To solve equation (1), we first multiply both sides of equation (1) by Pdxe (called integrating factor) and get

Pdx Pdx Pdxdye Pye Qedx

or Pdx Pdxd ye Qedx

.....(2)

Pdx Pdx Pdxd dyye e Py.e

dx dx∵

MATHEMATICS 463

Notes

MODULE - VCalculus


On integrating, we getPdx Pdxye Qe dx C .....(3)

where C is an arbitrary constant,

orPdx Pdxy e Qe dx C

Note : Pdxe is called the integrating factor of the equation and is written as I.F in short.

Remarks

(i) We observe that the left hand side of the linear differential equation (1) has become

Pdxd yedx

after the equation has been multiplied by the factor Pdxe .

(ii) The solution of the linear differential equationdy Py Qdx

P and Q being functions of x only is given byPdx Pdxye Q e dx C

(iii) The coefficient of dydx

, if not unity, must be made unity by dividing the equation by it

throughout.(iv) Some differential equations become linear differential equations if y is treated as the

independent variable and x is treated as the dependent variable.

For example, dx Px Qdy

, where P and Q are functions of y only, is also a linear

differential equation of the first order.

In this case PdyI.F. eand the solution is given by

x I.F. Q. I.F. dy C

Example 28.15 Solve

xdy y edx x

Solution : Here x1P , Q ex

(Note that both P an Q are functions of x)

I.F. (Integrating Factor)1dxPdx logxxe e e x x 0

On multiplying both sides of the equation by I.F., we get

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Notes

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464


xdyx y x edx

or xd y x xedx

Integrating both sides, we have

xyx xe dx C

where C is an arbitrary constant

or x xxy xe e dx C

or x xxy xe e C

or xxy e x 1 C

or xx 1 Cy e

x x

Note: In the solution x > 0.

Example 28.16 Solve :

2dysinx ycosx 2sin x cosxdx


2dysinx ycosx 2sin x cosxdx

ordy y c o t x 2 s i n x c o s xdx

.....(1)

Here P cotx,Q 2 s i n x c o s x

Pdx cotxdx logsinxI.F. e e e s inxOn multiplying both sides of equation (1) by I.F., we get (sin x >0)

2d ys inx 2sin x c o s xdx

Further on integrating both sides, we have2ysinx 2sin x cos x d x C

where C is an arbitrary constant (sin x >0)

or 32ysinx sin x C3

, which is the required solution.

Example 28.17 Solve 2 1dx1 y tan y xdy


MATHEMATICS 465

Notes

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2 1dx1 y tan y xdy

or1

2 2dx tan y xdy 1 y 1 y

or1

2 2dx x tan ydy 1 y 1 y ....(1)

which is of the form dx Px Qdy

, where P and Q are the functions of y only..

1 dy 12Pdy tan y1 yI.F. e e eMultiplying both sides of equation (1) by I.F., we get

11 1tan y tan y2

d tan yxe e

dy 1 yOn integrating both sides, we get

or 1tan y te x e .tdt C ,

where C is an arbitrary constant and 12

1t tan yanddt dy

1 y

or 1tan y t te x te e C

or 1tan y t te x te e C

or 1 1 1tan y 1 tan y tan ye x tan y e e C (on putting 1t tan y )

or 11 tan yx tan y 1 Ce


1. (i) Is2

2d y

y sin x a solution of y 0 ?dx

(ii) Is 3 dyy x a solution of x 4y 0 ?dx

2. Given below are some solutions of the differential equationdy 3xdx

.

State which are particular solutions and which are general solutions.

MATHEMATICS

Notes

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466


(i) 22y 3x (ii) 23y x 22

(iii) 22y 3x C (iv) 23y x 32

3. State whether the following differential equations are homogeneous or not ?

(i)2

2dy xdx 1 y (ii) 2 23xy y dx x xy dy 0

(iii) 2dyx 2 x 4x 9dx

(iv) 3 2 3 3x yx dy y x dx 0

4. (a) Show that y = a sin 2x is a solution of2

2d y

4y 0dx

(b) Verify that 3

3 23

d yy x ax c is a solution of 6

dx5. The general solution of the differential equation

2dy sec x is y tan x Cdx

.

Find the particular solution when

(a) x , y 14

(b) 2x , y 03

6. Solve the following differential equations :

(a) 5 1 3dy x tan xdx

(b) 3 2 xdy sin x cos x xedx

(c) 2 dy1 x xdx

(d) 2dy x sin3xdx

7. Find the particular solution of the equation x dye 4dx

, given that y = 3 , when x = 0


(a) 2 2 2 2dyx yx y xy 0dx

(b)dy xy x y 1dx

(c) 2 2sec x tanydx sec ytanxdy 0 (d) x y y 2dy e e xdx

9. Find the particular solution of the differential equationdy

log 3x 4y,dx given that y = 0 when x = 0


(a) 2 2x y dx 2xydy 0 (b)2dy yx y

dx x

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Notes

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(c)2 2dy x y y

dx x(d)

dy y ysindx x x

11. Solve :dy ysecx tan xdx


(a) 2 1dy1 x y tan xdx

(b) 2 dycos x y t a n xdx

(c)dyxlogx y 2logxdx

, x > 1

13. Solve the following differential equations:

(a)dyx y 1 1dx

[Hint:dx x y 1dy or

dx x y 1dy which is of the form

dx Px Qdy ]

(b) 2 dyx 2y ydx

, y > 0 [Hint: 2dxy x 2ydy or

dx x 2ydy y ]

28.7 DIFFERENTIAL EQUATIONS OF HIGHER ORDERTill now we were dealing with the differential equations of first order. In this section, simpledifferential equations of second order and third order will be discussed.

28.7.1 Differential Equations of the Type 2

2d y

f(x)dx

Consider the differential equation2

2d y

f(x)dx

It may be noted that it is a differential equation of second order. So its general solution willcontain two arbitrary constants.

Now we have,2

2d y

f(x)dx

....(1)

ord dy

f(x)dx dx

Integrating both sides of (1) , we get

1dy f x dx Cdx

, where 1C is an arbitrary constant

Let f x dx x

468 MATHEMATICS

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Notes


Then 1dy

(x) Cdx

= φ + ...(2)

Again on integrating both sides of (2), we get

1 2y (x)dx C x C ,= φ + +∫where C

2 is another arbitrary constant. Therefore in order to find the particular solution we

need two conditions. [See Example 28.19]

Example 28.18 Solve2

x2

d yxe

dx=


2x

2

d yxe

dx= ...(1)

Now integrating both sides of (1), we have

x12

dyxe dx C ,

dx= +∫ ...(1)

where C1is an arbitrary constant

orx x

1dy

xe e dx Cdx

= − +∫

orx x

1dy

xe e Cdx

= − + ...(2)

Again on integrating both sides of (2), we get

( )x x1 2y xe e C dx C ,= − + +∫

where C2 is another arbitrary constant.

or x x x1 2y xe e dx e C x C= − − + +∫

or x x1 2y xe 2e C x C= − + +

which is the required general solution.

Example 28.19 Find theparticular solution of the different equation

22

2

d yx sin 3x

dx= +

for which y(0)=0 and dy

0dx

= when x = 0


22

2

d yx sin 3x

dx= +

MATHEMATICS 469


MODULE - VCalculus

Notes

Integrating both sides of (1), we have

( )21

dyx sin 3x dx C ,

dx= + +∫

where C1 is an arbitrary constant

or3

1dy x cos3x

Cdx 3 3

= − + ...(2)

or3

1x cos3x

dy C dx3 3

= − +

Further on integrating both sides of (2), we get

or3

1 2x cos3x

y C dx C ,3 3

= − + +

∫ ...(3)

Now substituting y = 0, when x = 0 in (3), we have, 0 = C2

Again substituting dy

0dx

= when x = 0 in (2) we have

1 11 1

0 C or C3 3

= − + =

Putting values of C1 and C

2 in (3), we get

4x 1 xy sin 3x

12 9 3= − +


28.7.2 Differential Equations of the Type 3

3

d yf (x)

dx=

Consider the differential equation

3

3

d yf (x)

dx=

This is a differential equation of order three. Its general solution will contain three arbitraryconstants. In order to find its particular solution we need three conditions [see Example 28.21].

For solving the differential equation of order three of the form

3

3

d yf (x)

dx= , integrate the equation three times.

470 MATHEMATICS

MODULE - VCalculus

Notes


Example 28.20 Solve 3

43

d yx

dx=


34

3

d yx

dx=

or 2

42

d d yx

dx dx

=

...(1)

On integrating both sides of (1), we will get

2 5

12

d y xC ,

5dx= +

where C1 is an arbitrary constat.

or 5

1d dy x

Cdy dx 5

= + ...(2)

Further on integrating both sides of (2), we have

6

1 2dy x

C x Cdx 30

= + + ...(3)


On integrating both sides of (3), we have

7 2

1 2 3x x

y C C x C ,7(30) 2

= + + +

where C3 is also an arbitrary constant.

or 27

12 3

C xxy C x C

210 2= + + +

Example 28.21 Find the particular solution of the differential equation

3x

3

d yxe

dx=

for which y = 1, dy

2dx

= , when x = 0

and2

2

d y0,

dx= when x = 1.


MATHEMATICS 471


MODULE - VCalculus

Notes

3x

3

d yxe

dx=


2x

12

d yxe dx C

dx= +∫

where C1 is an arbitrary constant.

or2

x x12

d yxe e dx C

dx= − +∫

2x x

12

d yxe e C

dx= − + ...(2)

Again on integrating both sides of (2), we will get

( )x x1 2

dyxe e C dx C ,

dx= − + +∫


orx x x

1 2dy

xe e dx e C x Cdx

= − − + +∫

orx x

1 2dy

xe 2e C x Cdx

= − + + ...(3)

Further on integrating both sides of (3), we have

2x x x

1 2 3x

y xe e 2e C C x C ,2

= − − + + +

where C3 is also an arbitrary constant.

or2

x x1 2 3

xy xe 3e C C x C

2= − + + +

Now we have y = 1,dy

2dx

= when x = 0 and 2

2

d y0,

dx= when x = 1

On substituting2

2

d y0,

dx= , x = 1 in equation (2), we get

0 = e1− e

1 + C

1

∴ C1 = 0

Now putting dy

2dx

= , when x = 0 in equation (3), we have

2 = 0 − 2e0 + C1 . 0 + C

2

⇒ 2 = −2 + C2 ⇒ C

2 = 4

472 MATHEMATICS

MODULE - VCalculus

Notes


Now putting y = 1, when x = 0 in equation (4), we have

1 = 0 − 3.e0 + C1 . 0 + C

2 . 0 + C

3

∴ C3 = 4

Further on substituting values of C1, C2 and C3 in equation (4), we will get

y = xex − 3ex + 4x + 4

which is the required solution.


1. Find the particular solution of the differential equation

22

2

d ysin x

dx= given that dy

0dx

= , when x ; y 02

π= = when x = 0.

2. Solve the differential equation

3

3

d yx sin x

dx= + given that y = 0,

2

2

dy d y1, 2

dx dx= − = when x = 0

3. Solved the following differential equations :

(a)2

2

d yx sin x

dx= (b)

2

2

d y(1 cos 2x) 1 cos 2x

dx− = +

(c)2

22

d ysin x

dx= (d)

2

2

d ycos x sin x

dx= −

4. Solve the following initial value problems :

(a) 2

2

d yx 1,

dx= given that y = 1, dy

0,dx

= when x = 1

(b)2

2

d y1 sin x,

dx= + given that x = 0, y = 0, dy

0dx

=

(c)2

2

d y2,

dx= x = 1 when y = 3 and x = -1 when y = 1

5. Solve the follwing differential equations :

(a)3

2 x3

d yx e

dx= + (b)

3x

3

d ye cos x

dx= + (c)

3

3

d y0

dx=

28.8 APPLICATIONS OF DIFFFERENTIAL EQUATIONS

Here we consider the applications of differential equations insome of the problems of Physics,Economics, Biology and other related fields.

MATHEMATICS 473


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Notes

28.8.1 Problems of Velogicty and Acceleration

We know that velocitty 'v' and acceleration 'a' of a body at any instant are given by

velocity = rate of change of displacement

∴d

v (s),dt

= where s is the displacement.

And acceleration = rate of change of velocity

da (v)

dt=

d d(s)

dt dt =

2

2

d(s)

dt=

Example 28.22A stone is dropped from the top of a tower 44.1 m high. Velocity 'v' at anyinstant is given by v = 9.8 m/s (metres per second)

If it is given that distance after one second is 4.9m

(i) find the distance after one second is 4.9 m

(ii) value of 't' when stone hits the ground.

Solution : We are given that v = 9.8 t

∴d

(s) 9.8 tdt

=

On integrating both sides, we get

ds 9.8t dt=∫ ∫or s = 4.9 t2 + C, where C is an arbitrary constant ...(1)

It is given that s = 4.9 m when t = 1 second. Using this condition we get,

C = 0

Substituting the value of C in (1), we will get

s = 4.9 t2 ...(2)

Putting t = 2 in (2), we get

s = 4.9 ×22 or s = 19.6 m

So it is clear that the distance of the stone in 2 secinds is 19.6 m.

Now when stone hits the ground the distance is 44.1m. Thus, s = 4.1

Putting s = 44.1 in (2), the following will be the outcome.

44.1 = 4.9 t2

474 MATHEMATICS

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Notes


or2 44.1

t4.9

= or t2 = 9 or t = 3

Hence the stone hits the ground after 3 seconds.

28.8.2 Population Growth Problems

Rate of growth of population is proportional to the population i.e. the growth rate dx

dt is

proportional to x where x = x(t) denotes the number of bacteria present at time t. This fact canbe represented by means of the differential equation

dx

dt = kx where k is positive constant of proportinality.

Example 28.23 In a certain culture of bacteria the rate of increase is proportional to thenumber present.

(a) If it is found that the number doubles in 4 hours, how may times is the number expectedat the end of 12 hours ?

(b) If there are 104 at the end of 3 hours and 4 ×104 at the end of 5 hours, how many werethere in the begining ?

Solution : Let x denote the number of bacteria at time t hours. Then,

dxkx

dt= , where k is an arbitrary constant

ordx

kdtx

= ....(1)

On integrating both sides, we get

dxkdt log C,

x= +∫ ∫

where C is an arbitrary constant.

or log x = kt + log C

Note that the number x is positive.

orx

log ktC

= orx

ktC

=

or x = Cekt ...(2)

Assuming that x = x0 at time t = 0, we get

x0 = C

Putting value of C in (2) we get

x = x0ekt

MATHEMATICS 475


MODULE - VCalculus

Notes

(a) It is given that when t = 4, x = 2x0 substituting these in (3), we get

2x0 = x

0e4k or e4k = 2

When t = 12

x = x0 e12k or x = x

0(e4k)3

or x = x0 (2)3 or x = 8x

0

Hence the growth of bacteria is 8 times the original number at the end of 12 hours.

(b) Now it is given that at t = 3, x = 104 and at t = 5, xu = 4 104

Substituting these values in (2), we have

104 = Ce3k ...(4)

and 4 × 104 = Ce5k ...(5)

From (4) and (5), we get

4 × Ce3k = Ce5k or 4 = e2k

or log 4 = 2k or1

k log 42

=

or k log 4 log 2= =

Substituting value of k in equation (4), we get

104 = Ce3 log 2 or 104 = 3log 2Ce

or 104 = Ce3 log 2 or 104 = 8C

or410

C8

=

Substituting values of C and k in equation (2), we get

4t(log 2)10

x e8

= ...(6)

Substituting t = 0 in equation (6), we will get the number of bacteria in beginning.

∴4

010x e

8= × or

410 1x

8

×=

or225 10

x2

×= or x = 25 × 50 = 1250

28.8.3 Newton's Law of Cooling Problems

Newton's law of cooling states that the rate at which the temperature T = T(t) changes in acooling body is proportional to the difference betweem the instantaneous temperature T of thebody and the constant temperature T

0 of the surrounding.

476 MATHEMATICS

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Notes


i.e. 0dT

k(T T )dt

= − − , where k > 0. Negative sign indicates that if T > T0,

dT

dt is negative,

that is the temperature decreases with time.

Example 28.24 A body whose temperature is initially 100oC is allowed to cool in air,whose temperature remains at a constant 20oC. It is given that after 10 minutes the body hascooled to 40oC. Find the temperature of the body after half an hour.

Solution : Let T denote the instantaneous temperature of body and t denote the time. Thenaccording to the Newton's Law of Cooling

dTk(T 20),

dt= − − where k > 0, is a constant.

dTk dt

T 20= −

− ...(1)

On integrating both sides of (1), we get dTk dt

T 20= −

−∫ ∫log (T − 20) = −kt + log C, where C is an arbitrary constant

or T − 20 = Ce−kt

or T = 20 + Ce−kt ...(2)

Now substituting t = 0, T = 100 in equation (2), we get

100 = 20 + C

or C = 80

On sbstituting value of C in equation (2), we get

T = 20 + 80e−kt ...(3)

It is given that when t = 10, T = 40 and on substituting these in (3), we get

40 = 20 + 80e−10k or 80e−10t = 20

or10k 1

e4

− = or

110k 1

e4

− =

Thus

t101

T 20 804

= +

Now after half an hour i.e. when t = 30, we have

31

T 20 804

= +

or1 1 1

T 20 804 4 4

= + × × ×

or T = 20 + 1.25 or T = 21.25oC

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Notes


1. The population of a certain country is known to increse at the rat of 10% per year. Howlong does it take for the population to double ?

2. A body at temperature 80oF is placed at time t = 0 in a medium, the temperature ofwhich is maintained at 50oF. At the end of 5 minutes, the body has cooled to a tempera-ture of 70oF. When will the temperature of the body be 60oF ?

3. A particle is moving in a straight line, the veocity v(m/s) at any instant is given by v = 4t2.It is given that the distance of the particle at the end of the article at the end of 1 secondis 2 metres. Find the distance of the particle at the end of 3 seconds.

4. The number of bacteria in a yeast culture grows at a rate which is proportinal to thenumber present. If the population of a colony of yeast bacteria triples in 1 hour, howmany may be expected at the end of 5 hours ?

Example 28.25 Verify if 1m sin xy e

−= is a solution of

( )2

2 22

d y dy1 x m y 0

dxdx− − − =

Solution : We have,1m sin xy e

−= ...(1)

Differentiating (1) w.r.t. x, we get1msin x

2 2

dy me my

dx 1 x 1 x

−

= =− −

or2 dy

1 x mydx

− =

Squaring both sides, we get

( )2

2 2 2dy1 x m y

dx − =

Differentiating both sides, we get

( )2 2

2 22

dy dy d y dy2x 2 1 x . 2m y

dx dx dxdx − + − =

or ( )2

2 22

dy d yx 1 x m y

dx dx− + − =

or ( )2

2 22

d y dy1 x x m y 0

dxdx− − − =

478 MATHEMATICS

MODULE - VCalculus

Notes


Hence1msin xy e

−= is the solutin of

22 2

2

d y dy(1 x ) x m y 0

dxdx− − − =

Example 28.26 Find the particular solution ofdydxe x 1,= + given that y = 2 when x = 0


dydxe x 1= + or

dylog(x 1)

dx= +

or dy = log(x + 1) dx ...(1)

On integrating both sides of equation (1), we get

y log (x 1) . 1dx C= + +∫where C is an arbitrary constant

orx

y x log(x 1) dx Cx 1

= + − ++∫

orx 1 1

y x log(x 1) dx Cx 1

+ −= + − ++∫

or y = x log(x+1) − [x − log (x + 1)] + C

or y = (x + 1) log (x + 1) − x + C ...(2)

Substituting y = 2 and x = 0 in equation (2), we get

2 = log 1 − 0 + C or C = 2

Putting the value of C in equation (2), we get

y = (x + 1) log(x + 1) − x + 2


Example 28.27 Find the equation of the curve represented by

(y − yx) dx + (x + xy) dy = 0

and passing through the point (1, 1).


(y - yx) dx + (x + xy) dy = 0

or (x + xy) dy = (yx − y) dx

or x(1 + y) dy = y (x − 1) dx

or(1 y) x 1

dy dxy x

+ −= ...(1)

MATHEMATICS 479


MODULE - VCalculus

Notes

Integrating both sides of equation (1), we get

1 y x 1dy dx

y x

+ − = ∫ ∫

or1 1

1 dy 1 dxy x

+ = − ∫ ∫

or log y + y = x − log x + C

Since the curve is passing through the point (1, 1), therefore,

substituting x = 1, y = 1 in equation (2), we get

1 = 1 + C

or C = 0

Thus, the equation of the required curve is

log y + y = x − log x

or log(xy) = x − y

Example 28.28 Solve 2x 4x

x x

dy 3e 3e

dx e e−+=+

Solution : We have 2x 4x

x x

dy 3e 3e

dx e e−+=+

or3x x x

x x

dy 3e (e e )

dx e e

−

−+=

+or

3xdy3e

dx=

or dy = 3e3x dx ...(1)

Integrating both sides of (1), we get

3xy 3e dx C= +∫where C is an arbitrary constant.

or3xe

y 3 C3

= + or y = e3x + C

which is required soluton.

Example 28.29 Find the solution of the following differential equation :

2 2dy dyy x a y x

dx dx − = +

satisfying x = a, y = a

480 MATHEMATICS

MODULE - VCalculus

Notes



2 2dy dyy x a y x

dx dx − = +

or2 2dy dy

y ay x axdx dx

− = + or ( ) ( )2 2 dyy ay x ax

dx− = +

or 2 2

dy dx

y ay x ax=

− + ordy dx

y(1 ay) x(1 ax)=

− +

or1 a 1 a

dy dxy 1 ay x 1 ax

+ = − − + ...(1)

Integrating both sides of (1), we get

1 a 1 ady dx C

y 1 ay x 1 ax

+ = − + − + ∫ ∫

where C is an arbitrary conswtant

or 1a a

log y log(1 ay) log x log(1 ax) log Ca a

− − = − + +

where C = log C1 (x, y > 0), (1 − ay) > 0, (1 + ax) > 0

or log y − log(1 − ay) = log x − log (1 + ax) + log C1

or1C xy

log log1 ay 1 ax

= − +

or y(1 + ax) = C1x(1 − ay) ...(2)

Now using the condition that y = a, where x = a in equation (2), we get

( ) ( )2 21a 1 a C a 1 a+ = − or

2

1 2

1 aC

1 a

+=−

Substituting value of C1 in equation (2), we get

( )( ) ( )( )2 2y 1 ax 1 a x 1 ay 1 a+ − = − +



1. (a) If y = tan−1x, prove that ( )2

22

d y dy1 x 2x 0

dxdx+ + =

(b) y = ex sin x, prove that 2

2

d y dy2 2y 0

dxdx− + =

MATHEMATICS 481


MODULE - VCalculus

Notes

2. Find the particular solution of the differential equation

dylog 3x 4y,

dx = +

given that y = 0 when x = 0

3. (a) Find the equation of the curve represented by

dyxy x y 1

dx= + + + and passing through thepoint (2, 0)

(b) Find the equation of the curve represented by

cos xdyy cot x 5e

dx= = and passing through the point , 2

2

π

4. Solve : 3x 5x

x x

dy 4e 4e

dx e e−+=+


(a) dx + xdy = e−y sec2 y dy

(b) ( )2 1dy1 x 4x 3cot x

dx−+ − =

(c) (1 + y) xy dy = (1 − x2) (1 − y) dx

LET US SUM UP

● A diffferential equation is an equation involving independent variable, dependentvariable and the derivatives of dependent variable (and differentials) with respect toindependent variable.

● The order of a differential equation is the order of the highest derivative occurring in it.

● The degree of a differential equation is the degree of thehighest derivative after it hasbeen freed from radicals and fractions as far as the derivatives are concerned.

● A differential equation in which the dependent variable and its differential coefficientsoccur only in the first degree and are not multiplied together is called a linear differentialequation.

● A linear differential equation is always of the first degree.

● A general solution ofa differential equation is that solution which ontains as many as thenumber of arbitrary constants as the order of the differential equation.

● A general solution becomes a particular solution when particular values of the arbitraryconstants are determined satisfying the given conditions.

● The solution of the diffferential equation of the type dy

f (x)dx

= is obtained by integrating

both sides.

● The solution of the differential equation of the type dy

f (x)dx

= g(y) is obtained after

separting the variables and integrating both sides.

482 MATHEMATICS

MODULE - VCalculus

Notes


● The differential equation M(x, y) dx + N(x, y) dy = 0 is called homogeneous if M(x, y)and N(x, y) are homogeneous and are the same degree.

● The solution of a homogeneous differential equation is obtained by substituting y = vx orx = vy ad then separating the variables.

● The solution of the first order linear equationdy

Py Qdx

+ = is

Pdx Pdxye Q e dx C, ∫ ∫= + ∫ where C is an arbitrary constant.

The expression Pdx

e∫ is called the integrating factor of the differential equation and iswritten as I.F. in short.

● The solution of a second order differential equation 2

2

d yf (x)

dx= is obtained simply by

integrating it twice with respect to x and its general solution contains two arbitrary constants.

SUPPORTIVE WEBSITES


● http://mathworld.wolfram.com

TERMINAL EXERCISE

1. Find the order and degree of the differential equation :

(a)

422

2

d y dyx 0

dxdx

2 + =

(b) xdx + ydy = 0

(c)4

4

d y dy4 4y 5cos3x

dxdx− + = (d)

dycos x

dx=

(e)2

22

d y dyx xy y

dxdx− = (f)

2

2

d yy 0

dx+ =

(g)2

dy dyy x a 1

dx dx = + +

(h)

32 22

2

dy d y1 a

dx dx

+ = 2. Find which of the following equations are linear and which are non-linear

(a)dy

cos xdx

= (b)2dy y

y log xdx x

+ =

MATHEMATICS 483


MODULE - VCalculus

Notes

(c)

3 222

2

d y dyx 0

dxdx

+ = (d)

dyx 4 x

dx− =

(e) dx + dy = 0

3. From the differential equation corresponding to y2 − 2ay + x2 = a2 by eliminating a.

4. Find the differential equation by eliminating a, b, c form

y = ax2 + bx + c. Write its order and degree.

5. How many constants are contained in the general solution of

(a) Second order differential equation.

(b) Differential equation of order three.

(c) Differential equation of order five.

6. Show that y = a cos(log x) + b sin (log x) is a solution of the differential equation

22

2

d y dyx x y 0

dxdx+ + =


(a)2 dy

sin x 3cos x 4dx

= + (b)x y 2 ydy

e x edx

− −= +

(c)dy cos x sin y

0dx cos y

+ = (d) dy + xydx = xdx

(e)mdy

y tan x x cos mxdx

+ = (f) ( )2 1dx1 y tan y x

dy−+ = −

(g)2

22

d ycos x

dx= (h)

21

2

d ysin x

dx−=

(i)2

x2

d ye sin x

dx= − (j)

3

3

d ycos3x sin 3x

dx= −

8. If dvg cos kv, g, , k

dt= α − α being constants. Find v in term of t if v = 0, when t = 0.

9. Solve the differential equation

2

2

d ylog x,

dx= given that y = 1 and

dy1,

dx= − when x = 1.

10. If the temperature of air is290 K and a substance cools from 370 K to 330 K in 10minutes, find when the temperature will be 295 K.

484 MATHEMATICS

MODULE - VCalculus

Notes


ANSWERS


1. Order is 1 and degree is 2.

2. (a) Order 2, degree 1 (b) Order 1, degree 2

(c) Order 1, degree 1 (d) Order 2, degree 2

3. (a) Non - linear (b) Linear

(c) Liinear (d) Non - Linear

4.

3 22 22

2

dy d y1 r

dx dx

+ =

5. (a)22

2

d y dy dyxy x y 0

dx dxdx + − =

(b) ( )2

2 2 2dy dyx 2y 4xy x 0

dx dx − − − =

(c)2

2

d y dy6y 0

dxdx+ − = (d)

dyy (x 3) 2

dx= − +

(e) ( )2 2 dyx y 2xy 0

dx− − =


1. (i) Yes (ii) No

2. (i), (ii) and (iv) are particular solutions (iii) is the general solution

3. (ii), (iv) are homogeneous

5. (a) y = tan x (b) y tan x 3= +

6. (a) 6 1 3 3 1 31 1 1y x tan (x ) x tan (x ) C

6 6 6− −= − + +

(b)5 3 x1 1

y cos x cos x (x 1)e C5 3

= − + − +

(c)21

y log x 1 C2

= + + (d)31 1

y x cos3x C3 3

= − +

7. y = −4e−x + 7

8. (a)x 1 1

log Cy x y

= + + (b)2x

log y 1 x C2

+ = + +

(c) tan x tan y = C (d)3

y x xe e C

3= + +

MATHEMATICS 485


MODULE - VCalculus

Notes

9. 3e−4y + 4e3x = 7

10. (a) x = C(x2 − y2) (b) x + cy = y log |x|

(c)1 y

sin log | x | Cx

− = + (d)

ytan Cx

2x=

11. y(sec x + tan x) = sec x + tan x − x + C

12. (a) y = tan−1 x − 1 + Ce−tan x (b) y = tan x − 1 + Ce−tan x

(c)C

y log xlog x

= +

13. (a) x = Cey − (y + 2) (b) x = y2 + Cy


1.21 1 1

y x x cos 2x4 4 8 8

π= − + − 2.4

2x 3y x x cos x 1

24 2= + − + −

3. (a) y = −x sin x − 2 cos x+C1x + C2 (b) y = C1x + C2 − log sin x − 21x

2

(c)2

1 21 1

y x cos 2x C x C4 8

= + + + (d) 1 2y cos x sin x C x C= − + + +

4. (a) y = x(logx − 1) + 2 (b)21

y x sin x x2

= − +

(c) y = x2 + x + 1

5. (a)5

x 21 2 3

xy e C x C x C

60= + + + + (b) y = ex − sin x + C1x2 + C2x + C3

(c) y = C1x2 + C2x + C3


1. t = 10 log 2 years 2. 13.55 Minutes

3.2

12 minutes3

4. 243 Times


2. 4e3x + 3e−4y = 7

3. (a)21

log(y 1) x x 42

+ = + − (b) y sin x + 5ecosx = 7

4.5x4

y e C5

= +

486 MATHEMATICS

MODULE - VCalculus

Notes


5. (a) x = e−y (C + tan y) (b) ( )22 13y 2log 1 x cot x C

2−= + − +

(c)2 2x y

log x 2log |1 y | 2y C2 2

+ − = − − +

TERMINAL EXERCISE

1. (a) order 2, degree 3 (e) Order 2, degree 1

(b) Order 1, degree 1 (f) Order 2, degree 1

(c) Order 4, degree 1 (g) Order 1, degree 2

(d) Order 1, degree 1 (h) Order 2, degree 1

2. (a), (d), (e) are linear; (b), (c) are non-linear

3. ( )2

2 2 2dy dyx 2y 4xy x 0

dx dx − − − =

4.3

3

d y0,

dx= Order 3, degree 1.

5. (a) Two (b) Three (c) Five

7. (a) y + 3 cosec x + 4 cot x = C (b)3

y x xe e C

3= + +

(c) sin y = Ce−sin x (d)2x

log(1 y) C2

− + =

(e)m 1x

y cos x Ccos xm 1

+= +

+(f) 11 tan yx tan y 1 Ce

−− −= − +

(g)2

1 2x cos 2x

y C x C4 8

= − + +

(h)2

1 21 2

x 1 3y sin x x 1 x C x C

4 4 4−

= + + − + +

(i) y = ex + sin x + C1x + C2 (j) 2

12 3

C xsin 3x cos3xy C x C

27 27 2= − − + + +

8.ktg cos

v (1 e )k

−α= −

9.2

2x log x 3 7y x

2 4 4= − +

10. 40 minutes.

STATISTICS

29 Measures of Dispersion30 Random Experiments and Events31 Probability

MODULE-V

MATHEMATICS 489

Notes

MODULE - VIStatistics

Measures of Dispersion

29

MEASURES OF DISPERSION

You have learnt various measures of central tendency. Measures of central tendency help us torepresent the entire mass of the data by a single value.Can the central tendency describe the data fully and adequately?In order to understand it, let us consider an example.The daily income of the workers in two factories are :Factory A : 35 45 50 65 70 90 100Factory B : 60 65 65 65 65 65 70Here we observe that in both the groups the mean of the data is the same, namely, 65(i) In group A, the observations are much more scattered from the mean.(ii) In group B, almost all the observations are concentrated around the mean.Certainly, the two groups differ even though they have the same mean.Thus, there arises a need to differentiate between the groups. We need some other measureswhich concern with the measure of scatteredness (or spread).To do this, we study what is known as measures of dispersion.


explain the meaning of dispersion through examples;define various measures of dispersion range, mean deviation, variance and standarddeviation;calculate mean deviation from the mean of raw and grouped data;calculate variance and standard deviation for raw and grouped data; andillustrate the properties of variance and standard deviation.

EXPECTED BACKGROUND KNOWLEDGEMean of grouped dataMedian of ungrouped data

MATHEMATICS

Notes


490


29.1 MEANING OF DISPERSIONTo explain the meaning of dispersion, let us consider an example.Two sections of 10 students each in class X in a certain school were given a common test inMathematics (40 maximum marks). The scores of the students are given below :

Section A : 6 9 11 13 15 21 23 28 29 35

Section B: 15 16 16 17 18 19 20 21 23 25The average score in section A is 19.The average score in section B is 19.Let us construct a dot diagram, on the same scale for section A and section B (see Fig. 29.1)The position of mean is marked by an arrow in the dot diagram.

Section A

Section B

Fig. 29.1

Clearly, the extent of spread or dispersion of the data is different in section A from that of B. Themeasurement of the scatter of the given data about the average is said to be a measure ofdispersion or scatter.

In this lesson, you will read about the following measures of dispersion :

(a) Range

(b) Mean deviation from mean

(c) Variance

(d) Standard deviation

29.2 DEFINITION OF VARIOUS MEASURES OF DISPERSION(a) Range : In the above cited example, we observe that(i) the scores of all the students in section A are ranging from 6 to 35;(ii) the scores of the students in section B are ranging from 15 to 25.The difference between the largest and the smallest scores in section A is 29 (35 6)The difference between the largest and smallest scores in section B is 10 (25 15).Thus, the difference between the largest and the smallest value of a data, is termed as the rangeof the distribution.

MATHEMATICS 491

Notes



(b) Mean Deviation from Mean : In Fig. 29.1, we note that the scores in section B clusteraround the mean while in section A the scores are spread away from the mean. Let ustake the deviation of each observation from the mean and add all such deviations. If thesum is 'large', the dispersion is 'large'. If, however, the sum is 'small' the dispersion issmall.

Let us find the sum of deviations from the mean, i.e., 19 for scores in section A.

Observations ix Deviations from mean ix x6 139 1011 813 615 421 +223 +428 +929 +1035 16190 0

Here, the sum is zero. It is neither 'large' nor 'small'. Is it a coincidence ?

Let us now find the sum of deviations from the mean, i.e., 19 for scores in section B.

Observations ( ix ) Deviations from mean ix x

15 416 316 317 218 119 020 121 223 425 6

190 0Again, the sum is zero. Certainly it is not a coincidence. In fact, we have proved earlier that thesum of the deviations taken from the mean is always zero for any set of data. Why is thesum always zero ?On close examination, we find that the signs of some deviations are positive and of some otherdeviations are negative. Perhaps, this is what makes their sum always zero. In both the cases,

MATHEMATICS

Notes


492


we get sum of deviations to be zero, so, we cannot draw any conclusion from the sum ofdeviations. But this can be avoided if we take only the absolute value of the deviations andthen take their sum.If we follow this method, we will obtain a measure (descriptor) called the mean deviation fromthe mean.

The mean deviation is the sum of the absolute values of the deviations from themean divided by the number of items, (i.e., the sum of the frequencies).

(c) Variance : In the above case, we took the absolute value of the deviations taken frommean to get rid of the negative sign of the deviations. Another method is to square thedeviations. Let us, therefore, square the deviations from the mean and then take theirsum. If we divide this sum by the number of observations (i.e., the sum of the frequen-cies), we obtain the average of deviations, which is called variance. Variance is usuallydenoted by 2 .

(d) Standard Deviation : If we take the positive square root of the variance, we obtain theroot mean square deviation or simply called standard deviation and is denoted by .

29.3 MEAN DEVIATION FROM MEAN OF RAW ANDGROUPED DATA

Mean Deviation from mean of raw data =

ni

i 1x x

N

Mean deviation from mean of grouped data

ni i

i 1f x x

N

wheren n

i i ii 1 i 1

1N f , x f x

N

The following steps are employed to calculate the mean deviation from mean.Step 1 : Make a column of deviation from the mean, namely ix x (In case of grouped data

take ix as the mid value of the class.)

Step 2 : Take absolute value of each deviation and write in the column headed ix x .For calculating the mean deviation from the mean of raw data use

Mean deviation of Mean =

ni

i 1x x

N

For grouped data proceed to step 3.Step 3 : Multiply each entry in step 2 by the corresponding frequency. We obtain i if x x

and write in the column headed i if x x .

MATHEMATICS 493

Notes



Step 4 : Find the sum of the column in step 3. We obtainn

i ii 1

f x x

Step 5 : Divide the sum obtained in step 4 by N.Now let us take few examples to explain the above steps.

Example 29.1 Find the mean deviation from the mean of the following data :

Size of items ix 4 6 8 10 12 14 16

Frequency if 2 5 5 3 2 1 4

Mean is 10

Solution :

ix if ix x ix x i if x x

4 2 5.7 5.7 11.46 4 3.7 3.7 14.88 5 1.7 1.7 8.5

10 3 0.3 0.3 0.912 2 2.3 2.3 4.614 1 4.3 4.3 4.316 4 6.3 6.3 25.2

21 69.7

Mean deviation from mean = i if x x2l

69.7 3.31921

Example 29.2 Calculate the mean deviation from mean of the following distribution :

Marks 0 10 10 20 20 30 30 40 40 50

No. of Students 5 8 15 16 6

Mean is 27 marks

Solution :Marks Class Marks ix if ix x ix x i if x x

0 10 5 5 22 22 11010 20 15 8 12 12 9620 30 25 15 2 2 3030 40 35 16 8 8 12840 50 45 6 18 18 108

Total 50 472

MATHEMATICS

Notes


494


Mean deviation from Mean = i if x xN

=472 Marks 9.4450

Marks

CHECK YOUR PROGRESS 29.11. The ages of 10 girls are given below :

3 5 7 8 9 10 12 14 17 18

What is the range ?

2. The weight of 10 students (in Kg) of class XII are given below :

45 49 55 43 52 40 62 47 61 58

What is the range ?

3. Find the mean deviation from mean of the data

45 55 63 76 67 84 75 48 62 65

Given mean = 64.

4. Calculate the mean deviation from mean of the following distribution.

Salary (in rupees) 20 30 30 40 40 50 50 60 60 70 70 80 80 90 90 100

No. of employees 4 6 8 12 7 6 4 3

Given mean = Rs. 57.2

5. Calculate the mean deviation for the following data of marks obtained by 40 students in atest

Marks obtained 20 30 40 50 60 70 80 90 100

No. of students 2 4 8 10 8 4 2 1 1

6. The data below presents the earnings of 50 workers of a factory

Earnings (in rupees) : 1200 1300 1400 1500 1600 1800 2000

No. of workers : 4 6 15 12 7 4 2

Find mean deviation.7. The distribution of weight of 100 students is given below :

Weight (in Kg) 50 55 55 60 60 65 65 70 70 75 75 80

No. of students 5 13 35 25 17 5

Calculate the mean deviation.

MATHEMATICS 495

Notes



8. The marks of 50 students in a particular test are :

Marks 20 30 30 40 40 50 50 60 60 70 70 80 80 90 90 100

No. of students 4 6 9 12 8 6 4 1

Find the mean deviation for the above data.

29.4 VARIANCE AND STANDARD DEVIATION OF RAW DATA

If there are n observations, 1 2 nx , x ....,x , then2 2 2

1 2 n2 x x x x ..... x xVariance

n

or

n2

i2 i 1

x x;

n where

ni

i 1x

xn

The standard deviation, denoted by , is the positive square root of 2 . Thus

n2

ii 1

x x

n

The following steps are employed to calculate the variance and hence the standard deviation ofraw data. The mean is assumed to have been calculated already.

Step 1 : Make a column of deviations from the mean, namely, ix x .

Step 2 (check) : Sum of deviations from mean must be zero, i.e.,n

ii 1

x x =0

Step 3: Square each deviation and write in the column headed 2ix x .

Step 4 : Find the sum of the column in step 3.

Step 5 : Divide the sum obtained in step 4 by the number of observations. We obtain 2 .

Step 6 : Take the positive square root of 2 . We obtain (Standard deviation).

Example 29.3 The daily sale of sugar in a certain grocery shop is given below :

Monday Tuesday Wednesday Thursday Friday Saturday

75 kg 120 kg 12 kg 50 kg 70.5 kg 140.5 kg

The average daily sale is 78 Kg. Calculate the variance and the standard deviation of the abovedata.

MATHEMATICS

Notes


496


Solution : x 78 kg (Given)

ix ix x 2ix x

75 _ 3 9120 42 176412 _ 66 435650 _ 28 784

70.5 _ 7.5 56.25140.5 62.5 3906.25

0 10875.50

Thus

2i

2 i 1x x

n10875.50

6= 1812.58 (approx.)

and = 42.57 (approx.)

Example 29.4 The marks of 10 students of section A in a test in English are given below :

7 10 12 13 15 20 21 28 29 35

Determine the variance and the standard deviation.

Solution : Here ix 190x 1910 10

ix ix x 2ix x

7 12 14410 9 8112 7 4913 6 3615 4 1620 +1 121 +2 428 +9 8129 +10 10035 +16 256

0 768

Thus 2 768 76.810

,

and 76.8 8.76(approx)

MATHEMATICS 497

Notes



CHECK YOUR PROGRESS 29.21. The salary of 10 employees (in rupees) in a factory (per day) is

50 60 65 70 80 45 75 90 95 100

Calculate the variance and standard deviation.

2. The marks of 10 students of class X in a test in English are given below :

9 10 15 16 18 20 25 30 32 35

Determine the variance and the standard deviation.

3. The data on relative humidity (in %) for the first ten days of a month in a city are givenbelow:

90 97 92 95 93 95 85 83 85 75

Calculate the variance and standard deviation for the above data.

4. Find the standard deviation for the data

4 6 8 10 12 14 16

5. Find the variance and the standard deviation for the data

4 7 9 10 11 13 16

6. Find the standard deviation for the data.

40 40 40 60 65 65 70 70 75 75 75 80 85 90 90 100

29.5 STANDARD DEVIATION AND VARIANCE OF RAW DATAAN ALTERNATE METHOD

If x is in decimals, taking deviations from x and squaring each deviation involves even moredecimals and the computation becomes tedious. We give below an alternative formula for com-puting 2 . In this formula, we by pass the calculation of x .

We know2n

i2

i 1

x xn

2 2nii

i 1

x 2x x xn

n n2

ii2i 1 i 1

x 2x xx

n nn

2i

2i 1x

xn

ixx

n∵

MATHEMATICS

Notes


498


i.e.

2n

ini 12

i2 i 1

xx

nn

And 2

The steps to be employed in calculation of 2 and, hence by this method are as follows :

Step 1 : Make a column of squares of observations i.e. 2ix .

Step 2 : Find the sum of the column in step 1. We obtainn

2i

i 1x

Step 3 : Substitute the values ofn

2i

i 1x , n and

n

ii 1

x in the above formula. We obtain 2 .

Step 4 : Take the positive sauare root of 2 . We obtain .

Example 29.5 We refer to Example 29.4 of this lesson and re-calculate the variance and

standard deviation by this method.

Solution :

ix 2ix

7 4910 10012 14413 16915 22520 40021 44128 78429 84135 1225

190 43782n

ini 12

i2 i 1

xx

nn

2190437810

10

MATHEMATICS 499

Notes



4378 361010

76810 76.8

and 76.8 8.76 (approx)

We observe that we get the same value of 2 and by either methods.

29.6 STANDARD DEVIATION AND VARIANCE OF GROUPEDDATA : METHOD - I

We are given k classes and their corresponding frequencies. We will denote the variance andthe standard deviation of grouped data by 2

g and g respectively. The formulae are givenbelow :

K2

i i2 i 1g

f x x

N,

K

ii 1

N f

and 2g g

The following steps are employed to calculate 2g and, hence g : (The mean is assumed to

have been calculated already).Step 1 : Make a column of class marks of the given classes, namely ix

Step 2 : Make a column of deviations of class marks from the mean, namely, ix x . Of

course the sum of these deviations need not be zero, since ix ' s are no more theoriginal observations.

Step 3 : Make a column of squares of deviations obtained in step 2, i.e., 2ix x and

write in the column headed by 2ix x .

Step 4 : Multiply each entry in step 3 by the corresponding frequency.

We obtain 2i if x x .

Step 5 : Find the sum of the column in step 4. We obtain k

2i i

i 1f x x

Step 6 : Divide the sum obtained in step 5 by N (total no. of frequencies). We obtain 2g .

Step 7 : 2g g

Example 29.6 In a study to test the effectiveness of a new variety of wheat, an experiment

was performed with 50 experimental fields and the following results were obtained :

MATHEMATICS

Notes


500


Yield per Hectare Number of Fields(in quintals)

31 35 236 40 341 45 846 50 1251 55 1656 60 561 65 266 70 2

The mean yield per hectare is 50 quintals. Determine the variance and the standard deviation ofthe above distribution.

Solution :

Yield per Hectare No. of Class ix x 2ix x 2

i if x x(in quintal) Fields Marks

31 35 2 33 17 289 57836 40 3 38 12 144 43241 45 8 43 7 49 39246 50 12 48 2 4 4851 55 16 53 +3 9 14456 60 5 58 +8 64 32061 65 2 63 +13 169 33866 70 2 68 +18 324 648

Total 50 2900

Thus

n2

i i2 i 1g

f x x

N2900 58

50 and g 58 7.61 (approx)

29.7 STANDARD DEVIATION AND VARIANCE OF GROUPEDDATA : METHOD - II

If x is not given or if x is in decimals in which case the calculations become rather tedious, weemploy the alternative formula for the calculation of 2

g as given below:2k

i iki 12

i i2 i 1g

f xf x

NN

,k

ii 1

N f

MATHEMATICS 501

Notes



and 2g g

The following steps are employed in calculating 2g , and, hence g by this method:

Step 1 : Make a column of class marks of the given classes, namely, ix .Step 2 : Find the product of each class mark with the corresponding frequency. Write the

product in the column i ix f .

Step 3 : Sum the entries obtained in step 2. We obtain k

i ii 1

f x .

Step 4 : Make a column of squares of the class marks of the given classes, namely, 2ix .

Step 5 : Find the product of each entry in step 4 with the corresponding frequency. We obtain2

i if x .

Step 6 : Find the sum of the entries obtained in step 5. We obtaink

2i i

i 1f x .

Step 7 : Substitute the values of k

2i i

i 1f x , N and

k

i ii 1

f x in the formula and obtain

2g .

Step 8 : 2g g .

Example 29.7 Determine the variance and standard deviation for the data given in Example

29.6 by this method.

Solution :

Yields per Hectare if ix i if x 2ix 2

i if x(in quintals)

31 35 2 33 66 1089 217836 40 3 38 114 1444 433241 45 8 43 344 1849 1479246 50 12 48 576 2304 2764851 55 16 53 848 2809 4494456 60 5 58 290 3364 1682061 65 2 63 126 3969 793866 77 2 68 136 4624 9248

Total 50 2500 127900

MATHEMATICS

Notes


502


Substituting the values of k

2i i

i 1f x , N and

k

i ii 1

f x in the formula, we obtain

2

2g

250012790050

502900

50= 58

and g 58= 7.61 (approx.)

Again, we observe that we get the same value of 2g , by either of the methods.

CHECK YOUR PROGRESS 29.31. In a study on effectiveness of a medicine over a group of patients, the following results were

obtained : Percentage of relief 0 20 20 40 40 60 60 80 80 100 No. of patients 10 10 25 15 40Find the variance and standard deviation.

2. In a study on ages of mothers at the first child birth in a village, the following data wereavailable :

Age (in years) 18 20 20 22 22 24 24 26 26 28 28 30 30 32at first child birthNo. of mothers 130 110 80 74 50 40 16

Find the variance and the standard deviation.3. The daily salaries of 30 workers are given below:

Daily salary 0 50 50 100 100 150 150 200 200 250 250 300(In Rs.)No. of workers 3 4 5 7 8 3

Find variance and standard deviation for the above data.

29.8 STANDARD DEVIATION AND VARIANCE: STEPDEVIATION METHOD

In Example 29.7, we have seen that the calculations were very complicated. In order to simplifythe calculations, we use another method called the step deviation method. In most of the frequencydistributions, we shall be concerned with the equal classes. Let us denote, the class size by h.

MATHEMATICS 503

Notes



Now we not only take the deviation of each class mark from the arbitrary chosen 'a' but alsodivide each deviation by h. Let

ii

x auh

.....(1)

Then i ix hu a .....(2)

We know that x hu a .....(3)Subtracting (3) from (2) , we get

i ix x h u u .....(4)

In (4) , squaring both sides and multiplying by if and summing over k, we get

k k2 22

i i i ii 1 i 1

f x x h f u u .....(5)

Dividing both sides of (5) by N, we getk

2i i k2

2i 1i i

i 1

f x xh f u u

N N

i.e. 2 2 2x uh .....(6)

where 2x is the variance of the original data and 2

u is the variance of the coded data or

coded variance. 2u can be calculated by using the formula which involves the mean, namely,,k

22u i i

i 1

1f u u

N ,k

ii 1

N f .....(7)

or by using the formula which does not involve the mean, namely,2k

i iki 12

i i2 i 1u

f uf u

NN

,k

ii 1

N f

Example 29.8 We refer to the Example 29.6 again and find the variance and standard

deviation using the coded variance.

Solution : Here h = 5 and let a = 48.

Yield per Hectare Number Class ii

x 48u5 i if u 2

iu 2i if u

(in quintal) of fields if marks ix

31 35 2 33 3 6 9 1836 40 3 38 2 6 4 12

MATHEMATICS

Notes


504


41 45 8 43 1 8 1 846 50 12 48 0 0 0 051 55 16 53 +1 16 1 1656 60 5 58 +2 10 4 2061 65 2 63 +3 6 9 1866 70 2 68 +4 8 16 32

Total 50 20 124

Thus

2k

i iki 12

i i2 i 1u

f uf u

NN

22012450

50124 8

50

2u

5825

Variance of the original data will be2 2 2x u

58h 25 5825

and x 58

7.61 (approx)

We, of course, get the same variance, and hence, standard deviation as before.

Example 29.9 Find the standard deviation for the following distribution giving wages of 230

persons.

Wages No. of persons Wages No. of persons(in Rs) (in Rs)

70 80 12 110 120 50

80 90 18 120 130 45

90 100 35 130 140 20

100 110 42 140 150 8

MATHEMATICS 505

Notes



Solution :

Wages No. of class ii

x 105u10

2iu i if u 2

i if u

(in Rs.) persons if mark ix70 80 12 75 3 9 36 10880 90 18 85 2 4 36 72

90 100 35 95 1 1 35 35100 110 42 105 0 0 0 0110 120 50 115 +1 1 50 50120 130 45 125 +2 4 90 180130 140 20 135 +3 9 60 180140 150 8 145 +4 16 32 128

Total 230 125 7532

2 2 2i i ii

1 1h f u f uN N

2753 125100230 230

100 3.27 0.29 = 298

298 17.3 (approx)

CHECK YOUR PROGRESS 29.41. The data written below gives the daily earnings of 400 workers of a flour mill.

Weekly earning ( in Rs.) No. of Workers

80 100 16100 120 20120 140 25140 160 40160 180 80180 200 65200 220 60220 240 35240 260 30260 280 20280 300 9

Calculate the variance and standard deviation using step deviation method.

MATHEMATICS

Notes


506


2. The data on ages of teachers working in a school of a city are given below:

Age (in years) 20 25 25 30 30 35 35 40

No. of teachers 25 110 75 120

Age (in years) 40 45 45 50 50 55 55 60

No. of teachers 100 90 50 30

Calculate the variance and standard deviation using step deviation method.

3. Calculate the variance and standard deviation using step deviation method of the follow-ing data :

Age (in years) 25 30 30 35 35 40

No. of persons 70 51 47

Age (in years) 40 50 45 50 50 55

No. of persons 31 29 22

29.9 PROPERTIES OF VARIANCE AND STANDARDDEVIATION

Property I : The variance is independent of change of origin.

To verify this property let us consider the example given below.

Example : 29.10 The marks of 10 students in a particular examination are as follows:

10 12 15 12 16 20 13 17 15 10Later, it was decided that 5 bonus marks will be awarded to each student. Compare the varianceand standard deviation in the two cases.

Solution : Case I

ix if i if x ix x 2ix x 2

i if x x

10 2 20 _ 4 16 3212 2 24 _ 2 4 813 1 13 _ 1 1 115 2 30 1 1 216 1 16 2 4 417 1 17 3 9 920 1 20 6 36 36

10 140 92

Here140x 1410

MATHEMATICS 507

Notes



Variance2

i if x x10

9210 9.2

Standard deviation 9.2 3.03

Case II (By adding 5 marks to each ix )


i if x x

15 2 30 4 16 3217 2 34 2 4 818 1 18 1 1 120 2 40 1 1 221 1 21 2 4 422 1 22 3 9 925 1 25 6 36 36

10 190 92190x 1910

Variance92 9.210

Standard deviation 9.2 3.03Thus, we see that there is no change in variance and standard deviation of the given data if theorigin is changed i.e., if a constant is added to each observation.

Property II : The variance is not independent of the change of scale.

Example 29.11 In the above example, if each observation is multiplied by 2, then discuss the

change in variance and standard deviation.Solution : In case-I of the above example , we have variance = 9.2, standard deviation = 3.03.Now, let us calculate the variance and the Standard deviation when each observation is multipliedby 2.


i if x x20 2 40 8 64 12824 2 48 4 16 3226 1 26 2 4 430 2 60 2 4 832 1 32 4 16 1634 1 34 6 36 3640 1 40 12 144 144

10 280 368

MATHEMATICS

Notes


508


280x 2810

Variance368 36.810

Standard deviation 36.8 6.06

Here we observe that, the variance is four times the original one and consequently the standarddeviation is doubled.In a similar way we can verify that if each observation is divided by a constant then the varianceof the new observations gets divided by the square of the same constant and consequently thestandard deviation of the new observations gets divided by the same constant.

Property III : Prove that the standard deviation is the least possible root mean square deviation.

Proof : Let x a d

By definition, we have22

i i1s f x aN

2i i

1 f x x x aN

2 2i i i

1 f x x 2 x x x a x aN

22

i i i i ix a1 2f x x x a f x x f

N N N2 20 d

The algebraic sum of deviations from the mean is zero

or 2 2 2s d

Clearly 2s will be least when d = 0 i.e., when a x .Hence the root mean square deviation is the least when deviations are measured from the meani.e., the standard deviation is the least possible root mean square deviation.

Property IV : The standard deviations of two sets containing 1n , and 2n numbers are 1and 2 respectively being measured from their respective means 1m and 2m . If the twosets are grouped together as one of 1 2n n numbers, then the standard deviation ofthis set, measured from its mean m is given by

2 2 22 1 1 2 2 1 21 22

1 2 1 2

n n n n m mn n n n

Example 29.12 The means of two samples of sizes 50 and 100 respectively are 54.1 and

50.3; the standard deviations are 8 and 7. Find the standard deviation of the sample of size 150

MATHEMATICS 509

Notes



by combining the two samples.Solution : Here we have

1 2 1 2n 50,n 100,m 54.1,m 50.3

1 28and 7

2 2 22 1 1 2 2 1 21 22

1 2 1 2

n n n n m mn n n n

22

50 64 100 49 50 100 54.1 50.3150 150

23200 4900 2 3.8150 9

57.21

7.56 (approx)

Example 29.13 Find the mean deviation (M.D) from the mean and the standard deviation

(S.D) of the A.P.a, a + d, a + 2 d,......,a + 2n.d

and prove that the latter is greater than the former.Solution : The number of items in the A.P. is (2n + 1)

x a nd

Mean deviation about the mean

2n

r 0

1a rd a nd

2n 1

1 .2 nd n 1 d ...... d2n 1

2 1 2 ..... n 1 n d2n 1

2n n 1.d

2n 1 2n n 1 d

2n 1 .....(1)

Now2n

22

r 0

1a rd a nd

2n 1

2 22 2 22d n n 1 .... 2 12n 1

MATHEMATICS

Notes


510


2 n n 1 2n 12d2n 1 6

2n n 1 d3

n n 1d.

3 .....(2)

We have further, (2) > (1)

ifn n 1 n n 1

d d3 2n 1

or if 22n 1 3n n 1

or if 2n n 1 0, which is true for n > 0Hence the result.

Example 29.14 Show that for any discrete distribution the standard deviation is not less than

the mean deviation from the mean.

Solution : We are required to show that

S.D. M.D. from mean

or 2 2S. D M.D. from mean

i.e.2

2i i i i

1 1f x x f x xN N

or2

2i i ii

1 1f d f dN N

, where i id x x

or 22i i iiN f d f d

or 22 21 2 1 1 2 2 1 1 2 2f f .... f d f d ...... f d f d .....

or 2 21 2 1 2 1 2 1 2f f d d ..... 2f f d d .....

or 21 2 1 2f f d d ..... 0

which is true being the sum of perfect squares.

MATHEMATICS 511

Notes



Range : The difference between the largest and the smallest value of the given data.

Mean deviation from mean

ni i

i 1f x x

N

wheren

ii 1

N f ,n

i ii 1

1x f x

N

Variance

n2

i2 i 1

x x

n[for raw data]

Standard derivation

n2

ii 1

x x

nVariance for grouped data

k2

i i2 i 1g

f x x

N, ix is the mid value of the class.

Also, 2 22x uh and

k2 2

u i ii 1

1f u u

N

k

ii 1

N f

or

2k2

i ik2 i 1

i i2 i 1

u

f uf u

NN

wherek

ii 1

N f

Standard deviation for grouped data 2g g

LET US SUM UP

MATHEMATICS

Notes


512



TERMINAL EXERCISE

1. Find the mean deviation for the following data of marks obtained (out of 100) by10 students in a test

55 45 63 76 67 84 75 48 62 65

2. The data below presents the earnings of 50 labourers of a factory

Earnings (in Rs.) 1200 1300 1400 1500 1600 1800

No. of Labourers 4 7 15 12 7 5

Calculate mean deviation.

3. The salary per day of 50 employees of a factory is given by the following data.

Salary (in Rs.) 20 30 30 40 40 50 50 60

No. of employees 4 6 8 12

Salary (in rupees) 60 70 70 80 80 90 90 100

No. of employees 7 6 4 3

Calculate mean deviation.

4. Find the batting average and mean deviation for the following data of scores of 50 inningsof a cricket player:

Run Scored 0 20 20 40 40 60 60 80

No. of Innings 6 10 12 18

Run scored 80 100 100 120

No. of innings 3 1

5. The marks of 10 students in test of Mathematics are given below:

6 10 12 13 15 20 24 28 30 32

Find the variance and standard deviation of the above data.

6. The following table gives the masses in grams to the nearest gram, of a sample of 10 eggs.

46 51 48 62 54 56 58 60 71 75

Calculate the standard deviation of the masses of this sample.


MATHEMATICS 513

Notes



7. The weekly income ( in rupees ) of 50 workers of a factory are given below:

Income 400 425 450 500 550 600 650

No of workers 5 7 9 12 7 6 4

Find the variance and standard deviation of the above data.

8. Find the variance and standard deviation for the following data:

Class 0 20 20 40 40 60 60 80 80 100

Frequency 7 8 25 15 45

9. Find the standard deviation of the distribution in which the values of x are 1,2,......, N.The frequency of each being one.

MATHEMATICS

Notes


514


ANSWERS

CHECK YOUR PROGRESS 29.11. 15 2. 22 3. 9.4 4. 15.44

5. 13.7 6. 136 7. 5.01 8. 14.4

CHECK YOUR PROGRESS 29.21. Variance = 311, Standard deviation = 17.63

2. Variance = 72.9, Standard deviation = 8.5


4. Standard deviation = 4


6. Standard deviation = 17.6

CHECK YOUR PROGRESS 29.31. Variance = 734.96, Standard deviation = 27.1


3. Variance = 5489 , Standard deviation = 74.09

CHECK YOUR PROGRESS 29.41. Variance = 2194, Standard deviation = 46.84

2. Variance = 86.5 , Standard deviation = 9.3

3. Variance = 67.08 , Standard deviation = 8.19

TERMINAL EXERCISE1. 9.4 2. 124.48

3. 15.44 4. 52, 19.8


6. 8.8

7. Variance = 5581.25,Standard deviation = 74.7

8. Variance = 840, Standard deviation = 28.9

9. Standard deviation 2N 112

MATHEMATICS 515

Notes


Random Experiments and Events

30

RANDOM EXPERIMENTS AND EVENTS

In day-to-day life we see that before commencement of a cricket match two captains go for atoss. Tossing of a coin is an activity and getting either a 'Head' or a "Tail' are two possibleoutcomes. (Assuming that the coin does not stand on the edge). If we throw a die (of course fairdie) the possible outcomes of this activity could be any one of its faces having numerals, namely1,2,3,4,5 and 6..... at the top face.An activity that yields a result or an outcome is called an experiment. Normally there are variety ofoutcomes of an experiment and it is a matter of chance as to which one of these occurs when anexperiment is performed. In this lesson, we propose to study various experiments and their outcomes.


explain the meaning of a random experiments and cite examples thereof;explain the role of chance in such random experiments;define a sample space corresponding to an experiment;write a sample space corresponding to a given experiment; anddifferentiate between various types of events such as equally likely, mutually exclusive,exhaustive, independent and dependent events.

EXPECTED BACKGROUND KNOWLEDGEBasic concepts of probability

30.1 RANDOM EXPERIMENTLet us consider the following activities :(i) Toss a coin and note the outcomes. There are two possible outcomes, either a head (H)

or a tail (T).

MATHEMATICS

Notes


516

Random Experiments and Events(ii) In throwing a fair die, there are six possible outcomes, that is, any one of the six faces

1,2,...... 6.... may come on top.

(iii) Toss two coins simultaneously and note down the possible outcomes. There are fourpossible outcomes, HH,HT,TH,TT.

(iv) Throw two dice and there are 36 possible outcomes which are represented as below :

i.e. outcomes are (1,1), (1,2), (1,3), (1,4), (1,5), (1,6)

2,1 , 2,2 ,...., 2 , 6: : :6,1 , 6,2 ,...., 6, 6

Each of the above mentioned activities fulfil the following two conditions.(a) The activity can be repeated number of times under identical conditions.(b) Outcome of an activity is not predictable beforehand, since the chance play a role and

each outcome has the same chance of being selection. Thus, due to the chance playing arole, an activity is(i) repeated under identical conditions, and(ii) whose outcome is not predictable beforehand is called a random experiment.

Example 30.1 Is drawing a card from well shuffled deck of cards, a random experiment ?

Solution :(a) The experiment can be repeated, as the deck of cards can be shuffled every time before

drawing a card.(b) Any of the 52 cards can be drawn and hence the outcome is not predictable beforehand.Hence, this is a random experiment.

Example 30.2 Selecting a chair from 100 chairs without preference is a random experiment.

Justify.

Solution :(a) The experiment can be repeated under identical conditions.

MATHEMATICS 517

Notes


Random Experiments and Events(b) As the selection of the chair is without preference, every chair has equal chances of selection.

Hence, the outcome is not predictable beforehand. Thus, it is a random experiment.Can you think of any other activities which are not random in nature.Let us consider some activities which are not random experiments.(i) Birth of Manish : Obviously this activity, that is, the birth of an individual is not repeatable

and hence is not a random experiment.(ii) Multiplying 4 and 8 on a calculator.Although this activity can be repeated under identical conditions, the outcome is always 32.Hence, the activity is not a random experiment.

30.2 SAMPLE SPACEWe throw a die once, what are possible outcomes ? Clearly, a die can fall with any of its faces atthe top. The number on each of the faces is, therefore, a possible outcome. We write the set S ofall possible outcomes as

S = {1, 2, 3, 4, 5, 6}

Again, if we toss a coin, the possible outcomes for this experiment are either a head or a tail. Wewrite the set S of all possible outcomes as

S = {H, T}.The set S associated with an experiment satisfying the following properties :(i) each element of S denotes a possible outcome of the experiment.(ii) any trial results in an outcome that corresponds to one and only one element of the set S

is called the sample space of the experiment and the elements are called sample points.Sample space is generally denoted by S.

Example 30.3 Write the sample space in two tosses of a coin.

Solution : Let H denote a head and T denote a tail in the experiment of tossing of a coin.

S = { (H, H), (H, T), (T, H), (T, T) }.

Note : If two coins are tossed simultaneously then the sample space S can be written as

S = { H H, H T, T H, T T}.

Example 30.4 Consider an experiment of rolling a fair die and then tossing a coin.

Write the sample space.

MATHEMATICS

Notes


518

Random Experiments and EventsSolution : In rolling a die possible outcomes are 1, 2, 3, 4, 5 and 6. On tossing a coin the possibleoutcomes are either a head or a tail. Let H (head) = 0 and T (tail) = 1.

S = {(1, 0), (1, 1), (2, 0), (2, 1),(3, 0), (3, 1),(4, 0), (4, 1), (5, 0), (5, 1), (6, 0), (6, 1)}

n(S) 6 2 12

Example 30.5 Suppose we take all the different families with exactly 3 children. The

experiment consists in asking them the sex (or genders) of the first, second and third chid. Writedown the sample space.

Solution : Let us write 'B' for boy and 'G' for girl and construct the following tree diagram.

The sample space isS = {BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG}

The advantage of writing the sample space in the above form is that a question such as "Was thesecond child a girl" ? or " How many families have first child a boy ?" and so forth can beanswered immediately.

n (S) = 2 × 2 × 2 = 8Example 30.6 Consider an experiment in which one die is green and the other is red. Whenthese two dice are rolled, what will be the sample space ?

MATHEMATICS 519

Notes


Random Experiments and EventsSolution : This experiment can be displayed in the form of a tree diagram, as shown below :

Let ig and jr denote, the number that comes up on the green die and red die respectively. Then

an out-come can be represented by an ordered pair i jg , r , where i and j can assume any ofthe values 1, 2, 3, 4, 5, 6.Thus, a sample space S for this experiment is the set

i jS { g , r : 1 i 6,1 j 6}.

Also, notice that the multiplication principle (principle of counting) shows that the number ofelements in S is 36, since there are 6 choices for g and 6 choices for r, and 6 × 6 = 36

n(S) 36

Example 30.7 Write the sample space for each of the following experiments :(i) A coin is tossed three times and the result at each toss is noted.(ii) From five players A, B, C, D and E , two players are selected for a match.(iii) Six seeds are sown and the number of seeds germinating is noted.(iv) A coin is tossed twice. If the second throw results in a head, a die is thrown, otherwise a

coin is tossed.Solution :(i) S = { TTT, TTH, THT, HTT, HHT, HTH, THH, HHH}

number of elements in the sample space is 2 × 2 × 2 = 8

MATHEMATICS

Notes


520

Random Experiments and Events(ii) S= { AB, AC, AD, AE, BC, BD, BE, CD, CE, DE}. Here n ( S ) = 10(iii) S = { 0, 1, 2, 3, 4, 5, 6}. Here n (S) = 7(iv) This experiment can be displayed in the form of a tree-diagram as shown below :

Thus S = {HH1, HH2, HH3, HH4, HH5, HH6, HTH, HTT, TH1, TH2,TH3,TH4,TH5, TH6, TTH, TTT}

i.e. there are 16 outcomes of this experiment.

30.3. DEFINITION OF VARIOUS TERMSEvent : Let us consider the example of tossing a coin. In this experiment, we may be interested

in 'getting a head'. Then the outcome 'head' is an event.In an experiment of throwing a die, our interest may be in, 'getting an even number'. Then theoutcomes 2, 4 or 6 constitute the event. We have seen that an experiment which, though repeatedunder identical conditions, does not give unique results but may result in any one of the severalpossible outcomes, which constitute the sample space.Some outcomes of the sample space satisfy a specified description, which we call an 'event'.We often use the capital letters A, B, C etc. to represent the events.

Example 30.8 Let E denote the experiment of tossing three coins at a time. List all possible

outcomes and the events that(i) the number of heads exceeds the number of tails.(ii) getting two heads.

Solution :

MATHEMATICS 521

Notes


Random Experiments and EventsThe sample space S is

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

= 1 2 3 4 5 6 7 8w , w , w , w , w , w , w , w (say)

If 1E is the event that the number of heads exceeds the number of tails, and 2E the event gettingtwo heads. Then

1 1 2 3 5E w , w , w , w

and 2 2 3 5E w , w , w

30.3.1 Equally Likely EventsOutcomes of a trial are said to be equally likely if taking into consideration all the relevantevidences there is no reason to expect one in preference to the other.

Examples :(i) In tossing an unbiased coin, getting head or tail are equally likely events.(ii) In throwing a fair die, all the six faces are equally likely to come.(iii) In drawing a card from a well shuffled deck of 52 cards, all the 52 cards are equally likely

to come.30.3.2 Mutually Exclusive Events

Events are said to be mutually exclusive if the happening of any one of the them preludes thehappening of all others, i.e., if no two or more of them can happen simultaneously in the sametrial.

Examples :(i) In throwing a die all the 6 faces numbered 1 to 6 are mutually exclusive. If any one of these

faces comes at the top, the possibility of others, in the same trial is ruled out.(ii) When two coins are tossed, the event that both should come up tails and the event that

there must be at least one head are mutually exclusive.Mathematically events are said to be mutually exclusive if their intersection is a null set (i.e.,empty) 30.3.3 Exhaustive Events If we have a collection of events with the property that no matter what the outcome of theexperiment, one of the events in the collection must occur, then we say that the events in thecollection are exhaustive events.For example, when a die is rolled, the event of getting an even number and the event of getting anodd number are exhaustive events. Or when two coins are tossed the event that at least onehead will come up and the event that at least one tail will come up are exhaustive events.Mathematically a collection of events is said to be exhaustive if the union of these events is thecomplete sample space.

30.3.4 Independent and Dependent EventsA set of events is said to be independent if the happening of any one of the events does not affectthe happening of others. If, on the other hand, the happening of any one of the events influence

MATHEMATICS

Notes


522

Random Experiments and Eventsthe happening of the other, the events are said to be dependent.Examples :(i) In tossing an unbiased coin the event of getting a head in the first toss is independent of

getting a head in the second, third and subsequent throws.(ii) If we draw a card from a pack of well shuffled cards and replace it before drawing the

second card, the result of the second draw is independent of the first draw. But, however,if the first card drawn is not replaced then the second card is dependent on the first draw(in the sense that it cannot be the card drawn the first time).

CHECK YOUR PROGRESS 30.11. Selecting a student from a school without preference is a random experiment. Justify.2. Adding two numbers on a calculator is not a random experiment. Justify.3. Write the sample space of tossing three coins at a time.4. Write the sample space of tossing a coin and a die.5. Two dice are thrown simultaneously, and we are interested to get six on top of each of

the die. Are the two events mutually exclusive or not ?6. Two dice are thrown simultaneously. The events A, B, C, D are as below :

A : Getting an even number on the first die.B : Getting an odd number on the first die.C : Getting the sum of the number on the dice < 7.D : Getting the sum of the number on the dice > 7.State whether the following statements are True or False.(i) A and B are mutually exclusive.(ii) A and B are mutually exclusive and exhaustive.(iii) A and C are mutually exclusive.(iv) C and D are mutually exclusive and exhaustive.

7. A ball is drawn at random from a box containing 6 red balls, 4 white balls and 5 blueballs. There will be how many sample points, in its sample space?

8. In a single rolling with two dice, write the sample space and its elements.9. Suppose we take all the different families with exactly 2 children. The experiment consists

in asking them the sex of the first and second child.Write down the sample space.

LET US SUM UP

An activity that yields a result or an outcome is called an experiment.An activity repeated number of times under identical conditions and outcome of activityis not predictable is called Random Experiment.

LET US SUM UP

MATHEMATICS 523

Notes


Random Experiments and EventsThe set of possible outcomes of a random experiment is called sample space and elementsof the set are called sample points.Some outcomes of the sample space satisfy a specified description, which is called anEvent.Events are said to be Equally likely, when we have no preference for one rather than theother.If happening of an event prevents the happening of another event, then they are calledMutually Exclusive Events.The total number of possible outcomes in any trial is known as Exhaustive Events.A set of events is said to be Independent events, if the happening of any one of the eventsdoes not effect the happening of other events, otherwise they are called dependent events.


TERMINAL EXERCISE

1. A tea set has four cups and saucers. If the cups are placed at random on the saucers,write the sample space.

2. If four coins are tossed, write the sample space.3. If n coins are tossed simultaneously, there will be how many sample points ?

[Hint : try for n = 1,2, 3,4, .....]4. In a single throw of two dice, how many sample points are there ?


MATHEMATICS

Notes


524

Random Experiments and Events

ANSWERS

CHECK YOUR PROGRESS 30.11. Both properties are satisfied 2. Outcome is predictable

3. S HHH,HHT,HTH,HTT,THH,THT,TTH,TTT

4. H1,H2,H3,H4,H5,H6,T1,T2,T3,T4,T4,T6 5. No.

6. (i) True (ii) True (ii) False (iv) True 7. 15

8. 1,1 , 1, 2 , 1,3 , 1, 4 , 1, 5 , 1 ,62,1 , 2,2 , 2,3 , 2,4 , 2,5 , 2 , 63,1 , 3,2 , 3,3 , 3,4 , 3,5 , 3 ,64,1 , 4,2 , 4,3 , 4,4 , 4,5 , 4 , 65,1 , 5,2 , 5,3 , 5,4 , 5,5 , 5,66,1 , 6,2 , 6,3 , 6,4 , 6,5 , 6,6

9. {MM, MF, FM, FF}

TERMINAL EXERCISE

1.1 1 1 2 1 3 1 4 2 1 2 2 2 3 2 4

3 1 3 2 3 3 3 4 4 1 4 2 4 3 4 4

C S , C S , C S , C S , C S , C S , C S , C S ,C S , C S , C S , C S , C S , C S , C S , C S

2. 42 16 , { HHHH, HHHT, HHTH, HTHH, HHTT, HTHT, HTTH, HTTT,,THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT}

3. n2 4. 26 36

MATHEMATICS 525

Notes


Probability

31

PROBABILITY

In our daily life, we often used phrases such as 'It may rain today', or 'India may win the match'or ' I may be selected for this post.' These phrases involve an element of uncertainty. How canwe measure this uncertainty ? A measure of this uncertainty is provided by a branch ofMathematics, called the theory of probability. Probability Theory is designed to measure thedegree of uncertainty regarding the happening of a given event. The dictionary meaning of probabilityis ' likely though not certain to occur. Thus, when a coin is tossed, a head is likely to occur but maynot occur. Similarly, when a die is thrown, it may or may not show the number 6.In this lesson, we shall discuss some basic concepts of probability, addition and multiplicationtheorem of probability and applications of probability in our day to day life.

OBJECTIVESAfter studying this lesson, you will be able to :� define probability of occurance of an event;� cite through examples that probability of occurance of an event is a non-negative fraction,

not greater than one;� use permutation and combinations in solving problems in probability;� state and establish the addition theorems on probability and the conditions under which

each holds;� generalize the addition theorem of probability for mutually exclusive events;� state the multiplication theorem on probability for any two events;� state and prove the multiplication theorem on compound probability for independent events;� solve problems on probability using addition and multiplication theorems;� define conditional probability of an event; and� solve problems involving conditional probability.

EXPECTED BACKGROUND KNOWLEDGE� Knowledge of random experiments and events.

MATHEMATICS

Notes


526

Probability

� The meaning of sample space.� A standard deck of playing cards consists of 52 cards divided into 4 suits of 13 cards

each : spades, hearts, diamonds, clubs and cards in each suit are - ace, king, queen,jack, 10, 9, 8, 7, 6, 5, 4, 3 and 2. Kings, Queens and Jacks are called face cards andthe other cards are called number cards.

31.1 EVENTS AND THEIR PROBABILITYIn the previous lesson, we have learnt whether an activity is a random experiment or not. Thestudy of probability always refers to random experiments. Hence, from now onwards, theword experiment will be used for a random experiment only. In the preceeding lesson, we havedefined different types of events such as equally likely, mutually exclusive, exhaustive, indepen-dent and dependent events and cited examples of the above mentioned events.Here we are interested in the chance that a particular event will occur, when an experiment isperformed. Let us consider some examples.What are the chances of getting a ' Head' in tossing an unbiased coin ? There are only twoequally likely outcomes, namely head and tail. In our day to day language, we say that the coinhas chance 1 in 2 of showing up a head. In technical language, we say that the probability of

getting a head is 12

.

Similarly, in the experiment of rolling a die, there are six equally likely outcomes 1, 2,3,4,5 or 6.The face with number '1' (say) has chance 1 in 6 of appearing on the top. Thus, we say that the

probability of getting 1 is 16

.

In the above experiment, suppose we are interested in finding the probability of getting evennumber on the top, when a die is rolled. Clearly, the numbers possible are 2, 4 and 6 and thechance of getting an even number is 3 in 6. Thus, we say that the probability of getting an even

number is 36

, i.e., 12

.

The above discussion suggests the following definition of probability.If an experiment with 'n' exhaustive, mutually exclusive and equally likely outcomes, m outcomesare favourable to the happening of an event A, the probability 'p' of happening of A is given by

Number of favourable outcomesP ATotal number of possible outcomes

mpn ....(i)

Since the number of cases favourable to the non-happening of the event A are n m− , theprobability 'q' that 'A' will not happen is given by

n m mq 1n n−

= = −

1 p= − [Using (i)]

∴ p q 1.+ =

MATHEMATICS 527

Notes


Probability

Obviously, p as well as q are non-negative and cannot exceed unity.

i.e., 0 p 1≤ ≤ , 0 q 1≤ ≤

Thus, the probability of occurrence of an event lies between 0 and 1[including 0 and 1].

Remarks1. Probability 'p' of the happening of an event is known as the probability of success and

the probability 'q' of the non-happening of the event as the probability of failure.2. Probability of an impossible event is 0 and that of a sure event is 1

if P (A) = 1, the event A is certainly going to happen andif P (A) = 0, the event is certainly not going to happen.

3. The number (m) of favourable outcomes to an event cannot be greater than the totalnumber of outcomes (n).

Let us consider some examples

Example 31.1 A die is rolled once. Find the probability of getting a 5.

Solution : There are six possible ways in which a die can fall, out of these only one is favourableto the event.

∴1P(5)6

= .

Example 31.2 A coin is tossed once. What is the probability of the coin coming up with head ?

Solution : The coin can come up either 'head' (H) or a tail (T). Thus, the total possible out-comes are two and one is favourable to the event.

So,1P ( H )2

=

Example 31. 3 A die is rolled once. What is the probability of getting a prime number ?

Solution : There are six possible outcomes in a single throw of a die. Out of these; 2, 3 and 5are the favourable cases.

∴ P ( Prime Number )3 16 2

= =

Example 31.4 A die is rolled once. What is the probability of the number '7' coming up ?

What is the probability of a number 'less than 7' coming up ?Solution : There are six possible outcomes in a single throw of a die and there is no face of thedie with mark 7.

∴ P ( number 7 ) 0 06

= =

[Note : That the probability of impossible event is zero]

MATHEMATICS

Notes


528

Probability

As every face of a die is marked with a number less than 7,

∴6P ( 7 ) 16

≤ = =

[Note : That the probability of an event that is certain to happen is 1]

Example 31.5 In a simultaneous toss of two coins, find the probability of

(i) getting 2 heads (ii) exactly 1 head

Solution : Here, the possible outcomes areHH, HT, TH, TT.

i.e., Total number of possible outcomes = 4.(i) Number of outcomes favourable to the event (2 heads) = 1 (i.e., HH).

∴ P ( 2 heads ) = 14

.

(ii) Now the event consisting of exactly one head has two favourable cases,namely HT and TH .

∴ P ( exactly one head ) 2 14 2

= = .

Example 31.6 In a single throw of two dice, what is the probability that the sum is 9?

Solution : The number of possible outcomes is 6 × 6 = 36. We write them as given below :

1,1 1,2 1,3 1,4 1,5 1,62,1 2,2 2,3 2,4 2,5 2,63,1 3,2 3,3 3,4 3,5 3,64,1 4,2 4,3 4,4 4,5 4,65,1 5,2 5,3 5,4 5,5 5,66,1 6,2 6,3 6,4 6,5 6,6

Now, how do we get a total of 9. We have :3 + 6 = 94 + 5 = 95 + 4 = 96 + 3 = 9

In other words, the outcomes (3, 6), (4, 5), (5, 4) and (6, 3) are favourable to the said event,i.e., the number of favourable outcomes is 4.

Hence, P (a total of 9 ) = 4 1

36 9=

Example 31.7 From a bag containing 10 red, 4 blue and 6 black balls, a ball is drawn at

random. What is the probability of drawing(i) a red ball ? (ii) a blue ball ? (iii) not a black ball ?

MATHEMATICS 529

Notes


Probability

Solution : There are 20 balls in all. So, the total number of possible outcomes is 20. (Randomdrawing of balls ensure equally likely outcomes)(i) Number of red balls = 10

∴ P (a red ball ) 10 120 2

= =

(ii) Number of blue balls = 4

∴ P ( a blue ball )4 1

20 5= =

(iii) Number of balls which are not black = 10 + 4= 14

∴ P ( not a black ball) 14 720 10

= =

Example 31.8 A card is drawn at random from a well shuffled deck of 52 cards. If A is the

event of getting a queen and B is the event of getting a card bearing a number greater than 4 butless than 10, find P(A) and P (B).

Solution : Well shuffled pack of cards ensures equally likely outcomes.∴ the total number of possible outcomes is 52.(i) There are 4 queens in a pack of cards.

∴4 1P ( A )

52 13= =

(ii) The cards bearing a number greater than 4 but less than 10 are 5,6, 7,8 and 9.Each card bearing any of the above number is of 4 suits diamond, spade, club or heart.Thus, the number of favourable outcomes = 5 × 4 = 20

∴20 552 13

= =

Example 31.9 What is the chance that a leap year, selected at random, will contain 53

Sundays?Solution : A leap year consists of 366 days consisting of 52 weeks and 2 extra days. These twoextra days can occur in the following possible ways.(i ) Sunday and Monday(ii) Monday and Tuesday(iii) Tuesday and Wednesday(iv) Wednesday and Thursday(v) Thursday and Friday(vi) Friday and Saturday(vii) Saturday and Sunday

MATHEMATICS

Notes


530

Probability

Out of the above seven possibilities, two outcomes,e.g., (i) and (vii), are favourable to the event

∴ P ( 53 Sundays ) = 27


1. A die is rolled once. Find the probability of getting 3.2. A coin is tossed once. What is the probability of getting the tail ?3. What is the probability of the die coming up with a number greater than 3 ?4. In a simultaneous toss of two coins, find the probability of getting ' at least' one tail.5. From a bag containing 15 red and 10 blue balls, a ball is drawn 'at random'. What is the

probability of drawing (i) a red ball ? (ii) a blue ball ?6. If two dice are thrown, what is the probability that the sum is (i) 6 ? (ii) 8? (iii) 10?

(iv) 12?7. If two dice are thrown, what is the probability that the sum of the numbers on the two

faces is divisible by 3 or by 4 ?8. If two dice are thrown, what is the probability that the sum of the numbers on the two

faces is greater than 10 ?9. What is the probability of getting a red card from a well shuffled deck of 52 cards ?10. If a card is selected from a well shuffled deck of 52 cards, what is the probability of

drawing(i) a spade ? (ii) a king ? (iii) a king of spade ?

11. A pair of dice are thrown. Find the probability of getting(i) a sum as a prime number(ii) a doublet, i.e., the same number on both dice(iii) a multiple of 2 on one die and a multiple of 3 on the other.

12. Three coins are tossed simultaneously. Find the probability of getting(i) no head (ii) at least one head (iii) all heads

31.2. CALCULATION OF PROBABILITY USING COMBINATORICS(PERMUTATIONS AND COMBINATIONS)

In the preceding section, we calculated the probability of an event by listing down all the possibleoutcomes and the outcomes favourable to the event. This is possible when the number ofoutcomes is small, otherwise it becomes difficult and time consuming process. In general, wedo not require the actual listing of the outcomes, but require only the total number of possibleoutcomes and the number of outcomes favourable to the event. In many cases, these can befound by applying the knowledge of permutations and combinations, which you have alreadystudied.

MATHEMATICS 531

Notes


Probability

Let us consider the following examples :

Example 31.10 A bag contains 3 red, 6 white and 7 blue balls. What is the probability that

two balls drawn are white and blue ?Solution : Total number of balls = 3 + 6 + 7 = 16

Now, out of 16 balls, 2 can be drawn in 162C ways.

∴ Exhaustive number of cases = 162

16 15C 1202

Out of 6 white balls, 1 ball can be drawn in 61C ways and out of 7 blue balls, one can be drawn

is71C ways. Since each of the former case is associated with each of the later case, therefore

total number of favourable cases are 6 71 1C C 6 7 42 .

∴ Required probability 42 7

120 20

RemarksWhen two or more balls are drawn from a bag containing several balls, there are two waysin which these balls can be drawn.

(i) Without replacement : The ball first drawn is not put back in the bag, when thesecond ball is drawn. The third ball is also drawn without putting back the balls drawnearlier and so on. Obviously, the case of drawing the balls without replacement is thesame as drawing them together.

(ii) With replacement : In this case, the ball drawn is put back in the bag before drawingthe next ball. Here the number of balls in the bag remains the same, every time a ball isdrawn.

In these types of problems, unless stated otherwise, we consider the problem of withoutreplacement.

Example 31.11 Find the probability of getting both red balls, when from a bag containing 5

red and 4 black balls, two balls are drawn,(i) with replacement.(ii) without replacement.Solution : (i) Total number of balls in the bag in both the draws = 5 + 4 = 9Hence, by fundamental principle of counting, the total number ofpossible outcomes = 9 × 9 = 81.Similarly, the number of favourable cases = 5 × 5 = 25.

Hence, probability (both red balls) =2581

.

MATHEMATICS

Notes


532

Probability

(ii) Total number of possible outcomes is equal to the number of ways of selecting 2 balls out

of 9 balls = 92C .

Number of favourable cases is equal to the number of ways of selecting

2 balls out of 5 red balls = 52C .

Hence, P (both red balls ) =5

29

2

5 4C 51 2

9 8 18C1 2

Example 31.12 Six cards are drawn at random from a pack of 52 cards. What is the

probability that 3 will be red and 3 black?

Solution : Six cards can be drawn from the pack of 52 cards in526C ways.

i.e., Total number of possible outcomes = 526C

3 red cards can be drawn in 263C ways and

3 black cards can be drawn in 263C ways.

∴ Total number of favourable cases = 263C × 26

3C

Hence, the required probability =26 26

3 352

6

C C 1300039151C

Example 31.13 Four persons are chosen at random from a group of 3 men, 2 women and

4 children. Show that the chance that exactly two of them will be children is 1021

.

Solution : Total number of persons in the group = 3 + 2 + 4 = 9. Four persons are chosen atrandom. If two of the chosen persons are children, then the remaining two can be chosen from5 persons (3 men + 2 women).Number of ways in which 2 children can be selected from 4

children = 42

4 3C 6

1 2×

= =×

Number of ways in which remaining of the two persons can be selected

from 5 persons = 52

5 4C 10

1 2×

= =×

Total number of ways in which 4 persons can be selected out of

9 persons = 94

9 8 7 6C 1261 2 3 4× × ×

= =× × ×

Hence, the required probability = 4 5

2 29

4

C CC

6 10 10126 21×

= =

MATHEMATICS 533

Notes


Probability

Example 31.14 Three cards are drawn from a well-shuffled pack of 52 cards. Find the

probability that they are a king, a queen and a jack.

Solution : From a pack of 52 cards, 3 cards can be drawn in 523C ways, all being equally

likely.∴ Exhaustive number of cases = 52

3C

A pack of cards contains 4 kings, 4 queens and 4 jacks .A king, a queen and a Jack can eachbe drawn in 4

1C ways and since each way of drawing a king can be associated with each of the

ways of drawing a queen and a jack, the total number of favourable cases = 41C × 4

1C 41C×

∴ Required probability 4 4 4

1 1 152

3

C C CC

4 4 452 51 50

1 2 3

165525

Example 31.15 From 25 tickets, marked with the first 25 numerals, one is drawn at

random. Find the probability that it is a multiple of 5.Solution : Numbers (out of the first 25 numerals ) which are multiples of 5 are 5, 10, 15, 20and 25, i.e., 5 in all. Hence, required favourable cases are = 5.

∴ Required probability = 5 125 5

Example 31.16 If the letters of the word 'REGULATIONS' be arranged at random, what

is the probability that there will be exactly 4 letters between R and E?Solution : The word 'REGULATIONS' consists of 11 letters. The two letters R and E canoccupy 11

2P , i.e., 11×10=110 positions.

The number of ways in which there will be exactly 4 letters between R and E are enumeratedbelow :

R is in the first place and E in the 6th place.R is in the 2nd place and E in the 7th place------- ------- ------- -------------- ------- ------- -------------- ------- ------- -------R is in the 6th place and E is in the 11th place.

Since R and E can interchange their positions, the requirednumber of favourable cases is 2 × 6 = 12.

MATHEMATICS

Notes


534

Probability

∴ The required probability 12 6

110 55

Example 31.17 Out of (2n + 1) tickets consecutively numbered starting with 1, three are

drawn at random. Find the chance that the numbers on them are in A.P.

Solution : Since out of (2n + 1 ) tickets, 3 tickets can be drawn in 2n 13C ways,

Therefore, exhaustive number of cases = 2n 13C

2n 1 2n 2n 11 2 3

2n 4n 13

To find the favourable number of cases, we are to enumerate all the cases in which the numberon the drawn tickets are in A.P. with common difference (say, d = 1, 2, 3, ......., n − 1, n)If d = 1, possible cases are as follows :

1, 2, 32, 3, 4

, i.e., 2n 1 cases in all.:2n 1, 2n, 2n 1

If d = 2, the possible cases are as follows :

1, 3, 52, 4, 6

, 2n 3 cases in all.:2n 3, 2n 1, 2n 1

and so onIf d = n − 1, the possible cases are

1, n, 2n 12, n 1, 2n , i.e., 3 cases in all.3, n 2, 2n 1

If d = n, there is only one case, viz., 1, n + 1, 2n + 1Hence, total number of favourable cases

= ( 2n− 1 ) + ( 2n− 3 ) +......+ 5 + 3 + 1= 1 + 3 + 5 + ....... + ( 2n − 1 )

which is a series in A.P. with d = 2 and n terms.

MATHEMATICS 535

Notes


Probability

∴ Number of favourable casesn 2 n 1 22

2n (2n) n2

Hence, required probability 2

22n 3n

4n 1n 4n 13

CHECK YOUR PROGRESS 31.21. A bag contains 3 red, 6 white and 7 blue balls. What is the probability that two balls

drawn at random are both white?

2. A bag contains 5 red and 8 blue balls. What is the probability that two balls drawn arered and blue ?

3. A bag contains 20 white and 30 black balls. Find the probability of getting 2 white balls,when two balls are drawn at random

(a) with replacement (b) without replacement

4. Three cards are drawn from a well-shuffled pack of 52 cards. Find the probability that allthe three cards are jacks.

5. Two cards are drawn from a well-shuffled pack of 52 cards. Show that the chances of

drawing both aces is1

221.

6. In a group of 10 outstanding students in a school, there are 6 boys and 4 girls. Threestudents are to be selected out of these at random for a debate competition. Find theprobability that

(i) one is boy and two are girls.

(ii) all are boys.

(iii) all are girls.

7. Out of 21 tickets marked with numbers from 1 to 21, three are drawn at random. Findthe probability that the numbers on them are in A.P.

8. Two cards are drawn at random from 8 cards numbered 1 to 8. What is the probabilitythat the sum of the numbers is odd, if the cards are drawn together ?

9. A team of 5 players is to be selected from a group of 6 boys and 8 girls. If the selectionis made randomly, find the probability that there are 2 boys and 3 girls in the team.

10. An integer is chosen at random from the first 200 positive integers.Find the probabilitythat the integer is divisible by 6 or 8.

MATHEMATICS

Notes


536

Probability

31.3 EVENT RELATIONSLet us consider the example of throwing a fair die. The sample space of this experiment is

S ={ 1 , 2, 3, 4, 5, 6 }If A be the event of getting an even number, then the sample points 2, 4 and 6 are favourable tothe event A.The remaining sample points 1, 3 and 5 are not favourable to the event A. Therefore, these willoccur when the event A will not occur.In an experiment, the outcomes which are not favourable to the event A are called complementof A and defined as follows :' The outcomes favourable to the complement of an event A consists of all those outcomes which arenot favourable to the event A, and are denoted by 'not' A or by A .

31.3.1 Event 'A or B'Let us consider the example of throwing a die. A is an event of getting a multiple of 2 and B beanother event of getting a multiple of 3.The outcomes 2, 4 and 6 are favourable to the event A and the outcomes 3 and 6 are favourableto the event B.

Fig. 31.1The happening of event A or B is

A B { 2, 3, 4, 6 }∪

Again, if A be the event of getting an even number and B is another event of getting an oddnumber, then

A = { 2, 4, 6 }, B = { 1, 3, 5 }

Fig. 31.2

A B {1, 2, 3, 4,5, 6 }∪

Here, it may be observed that if A and B are two events, then the event ' A or B ' ( A B∪ ) willconsist of the outcomes which are either favourable to the event A or to the event B or to boththe events.Thus, the event 'A or B' occurs, if either A or B or both occur.

MATHEMATICS 537

Notes


Probability

31.3.2 Event "A and B'Recall the example of throwing a die in which A is the event of getting a multiple of 2 and B is theevent of getting a multiple of 3. The outcomes favourable to A are 2, 4, 6 and the outcomesfavourable to B are 3 , 6.

Fig. 31.3

Here, we observe that the outcome 6 is favourable to both the events A and B.Draw a card from a well shuffled deck of 52 cards. A and B are two events defined as

A : a red cardB : a king

We know that there are 26 red cards and 4 kings in a deck of cards. Out of these 4 kings, twoare red.

Fig. 31.4

Here, we see that the two red kings are favourable to both the events.Hence, the event ' A and B' consists of all those outcomes which are favourable to both theevents A and B. That is, the event 'A and B' occurs, when both the events A and B occursimultaneously. Symbolically, it is denoted as A B.∩

31.4 ADDITIVE LAW OF PROBABILITYLet A be the event of getting an odd number and B be the event of getting a prime number in asingle throw of a die. What will be the probability that it is either an odd number or a primenumber ?In a single throw of a die, the sample space would be

S = { 1, 2, 3, 4, 5, 6 }

The outcomes favourable to the events A and B areA = { 1, 3, 5 }

MATHEMATICS

Notes


538

Probability

B = { 2, 3, 5 }

Fig. 31.5

The outcomes favourable to the event 'A or B' areA B∪ = {1, 2, 3, 5 }.

Thus, the probability of getting either an odd number or a prime number will be4 2P(A or B)6 3

To discover an alternate method, we can proceed as follows :The outcomes favourable to the event A are 1, 3 and 5.

∴3P ( A )6

Similarly,3P ( B )6

The outcomes favourable to the event 'A and B' are 3 and 5.

∴2P (A and B )6

Now, 3 3 2P(A) P(B) P(A and B)6 6 6

4 26 3

= P ( A or B )Thus, we state the following law, called additive rule, which provides a technique for finding theprobability of the union of two events, when they are not disjoint.For any two events A and B of a sample space S,

P ( A or B ) = P (A) + P (B) P ( A and B )

or ( )P A B P ( A ) P ( B ) P ( A B)= + −∪ ∩ .....(ii)

Example 31.18 A card is drawn from a well-shuffled deck of 52 cards. What is the probability that it is either a spade or a king ?Solution : If a card is drawn at random from a well-shuffled deck of cards, the likelyhood ofany of the 52 cards being drawn is the same. Obviously, the sample space consists of 52sample points.

MATHEMATICS 539

Notes


Probability

If A and B denote the events of drawing a 'spade card' and a 'king' respectively, then the eventA consists of 13 sample points, whereas the event B consists of 4 sample points. Therefore,

13P ( A )52

,4P ( B )

52

The compound event A B∩ consists of only one sample point, viz.; king of spade. So,

1P A B52

∩

Hence, the probability that the card drawn is either a spade or a king is given byP A B P ( A ) P ( B ) P ( A B )∪ ∩

13 4 152 52 521652

413

Example 31.19 In an experiment with throwing 2 fair dice, consider the events

A : The sum of numbers on the faces is 8B : Doubles are thrown.

What is the probability of getting A or B ?Solution : In a throw of two dice, the sample space consists of 6 × 6 = 36 sample points.The favourable outcomes to the event A ( the sum of the numbers on the faces is 8 ) are

A= { ( 2, 6 ) , ( 3, 5 ), ( 4, 4 ), ( 5, 3 ), ( 6, 2 ) }The favourable outcomes to the event B (Double means both dice have the same number) are

B 1,1 , 2,2 , 3,3 , 4,4 , 5,5 , 6 ,6

A B { (4,4) }.∩

Now 5P ( A )36

, 6P ( B )36

, 1P A B36

∩

Thus, the probability of A or B is5 6 1P A B

36 36 36∪

1036

518

31.5 ADDITIVE LAW OF PROBABILITY FORMUTUALLY EXCLUSIVE EVENTS

We know that the events A and B are mutually exclusive, if and only if they have no outcomesin common. That is, for mutually exclusive events,

P ( A and B ) = 0

MATHEMATICS

Notes


540

Probability

Substituting this value in the additive law of probability, we get the following law : P ( A or B ) = P ( A ) + P ( B ) .....(iii)

Example 31.20 In a single throw of two dice, find the probability of a total of 9 or 11.

Solution : Clearly, the events - a total of 9 and a total of 11 are mutually exclusive.

Now P ( a total of 9 ) = P [(3, 6), (4, 5), (5, 4), (6, 3)] = 4

36

P ( a total of 11 ) = P [ (5, 6), (6, 5)] =2

36

Thus, P ( a total of 9 or 11 ) = 4 2

36 36

16

Example 31.21 The probabilities that a student will receive an A, B , C or D grade are 0.30,

0.35, 0.20 and 0.15 respectively. What is the probability that a student will receive at least a Bgrade ?Solution : The event at least a 'B' grade means that the student gets either a B grade or an Agrade.∴ P ( at least B grade ) = P ( B grade ) + P ( A grade )

= 0.35 + 0.30= 0.65

Example 31.22 Prove that the probability of the non-occurrence of an event A is 1 P ( A).

i.e., P ( not A ) = 1 P ( A ) or, P ( A ) = 1 P ( A ).

Solution : We know that the probability of the sample space S in any experiment is 1.

Now, it is clear that if in an experiment an event A occurs, then the event A cannot occursimultaneously, i.e., the two events are mutually exclusive.Also, the sample points of the two mutually exclusive events together constitute the samplespace S. That is,

A A S∪

Thus, P A A P(S)∪

P A P ( A ) 1 (∵ A and A are mutually exclusive and S is sample space)

P A 1 P ( A ) ,

which proves the result.This is called the law of complementation.

Law of complimentation : P A 1 P ( A )

MATHEMATICS 541

Notes


Probability

Example 31.23 Find the probability of the event getting at least 1 tail, if four coins are tossed

once.Solution : In tossing of 4 coins once, the sample space has 16 samples points.

P ( at least one tail ) = P ( 1or 2 or 3 or 4 tails ) = 1 P ( 0 tail ) (By law of complimentation) = 1 P ( H H H H )

The outcome favourable to the event four heads is 1.

P ( H H H H ) = 1

16Substituting this value in the above equation, we get

P ( at least one tail ) = 11

16=

1516

In many instances, the probability of an event may be expressed as odds - either odds in favourof an event or odds against an event. If A is an event :

The odds in favour of A = P(A) or P ( A ) to P ( A ) ,P(A)

where P (A) is the probability of the event A and P ( A ) is the probability of the event 'not A'.

Similarly, the odds against A are

P(A) or P ( A ) to P ( A )P(A) .

Example 31.24 The probability of the event that it will rain is 0.3. Find the odds in favour of

rain and odds against rain.Solution : Let A be the event that it will rain.

P ( A ) = .3By law of complementation,

P A 1 .3 .7 .

Now, the odds in favour of rain are 0.30.7

or 3 to 7 (or 3 : 7).

The odds against rain are

0.7 or 7 to 3.0.3

When either the odds in favour of A or the odds against A are given, we can obtain the probabilityof that event by using the following formulae

MATHEMATICS

Notes


542

Probability

If the odds in favour of A are a to b, thenaP (A )

a b .

If the odds against A are a to b, then

bP (A )a b .

This can be proved very easily.Suppose the odds in favour of A are a to b. Then, by the definition of odds,

P(A) aP(A) b .

From the law of complimentation,

P A 1 P (A)

Therefore,P(A) a

1 P ( A ) b or b P (A) a a P (A)

or ( a b ) P ( A ) a oraP ( A )

a bSimilarly, we can prove that

bP (A)a b

when the odds against A are b to a.

Example 31.25 Determine the probability of A for the given odds

(a) 3 to 1 in favour of A (b) 7 to 5 against A.

Solution : (a)3 3P (A)

3 1 4(b)

5 5P (A)7 5 12

Example 31.26 If two dice are thrown, what is the probability that the sum is

(a) greater than 8 ? (b) neither 7 nor 11 ?

Solution : (a) If S denotes the sum on two dice, then we want P(S 8 ) . The required eventcan happen in the following mutually exclusive ways :

(i) S = 9 (ii) S = 10, (iii) S = 11, and ( iv) S = 12 .Hence, by addition probability theorem for mutually exclusive events, we get

P(S 8) P(S 9) P(S 10) P(S 11) P(S 12) ....(1)

In a throw of two dice, the sample space contains 6 × 6 = 36 points. The number of favourablecases can be enumerated as shown below :S = 9 : ( 3, 6 ), ( 4, 5 ), ( 5, 4 ), ( 6, 3) , i.e., 4 sample points.

MATHEMATICS 543

Notes


Probability

P ( S = 9 ) = 4

36.

S = 10 : ( 4, 6 ), ( 5, 5 ), ( 6, 4 ), , i.e., 3 sample points.

P ( S =10 ) = 3

36.

S = 11 : ( 5, 6 ), ( 6, 5 ), i.e., 2 sample points.

P ( S =11 ) = 2

36.

S = 12 : ( 6, 6 ), i.e., 1 sample point.

P ( S =12 ) = 1

36.

Substituting these values in equation (1), we get4 3 2 1 10 5P(S 8) .

36 36 36 36 36 18

(b) Let A and B denote the events of getting the sum 7 and 11 respectively on a pair of dice.

S = 7 : ( 1, 6 ), ( 2, 5 ), ( 3, 4 ), ( 4, 3 ) ,( 5 , 2 ), ( 6, 1), i.e., 6 sample points.

P ( S =7 ) = 6

36, or P ( A) =

636

.

S = 11 : ( 5, 6 ) , ( 6, 5 ), i.e., 2 sample points.2 2P ( S 11 ) , or P ( B )

36 36.

Since A and B are disjoint events, thereforeP(eitherAorB) P(A) P(B)

6 236 36

836

Hence, by law of complementation,P (neither 7 nor 11 ) = 1 P ( either 7 or 11)

8136

2836

79

.

Example 31.27 Are the following probability assignments consistent ? Justify your answer.

MATHEMATICS

Notes


544

Probability

(a) P (A) = P ( B ) = 0.6, P ( A and B ) = 0.05(b) P ( A) = 0.5, P ( B ) = 0.4, P ( A and B ) = 0.1(c) P ( A ) = 0.2, P ( B ) = 0.7, P ( A and B ) = 0.4Solution : (a) P ( A or B ) = P (A) + P ( B ) P ( A and B )

= 0.6 + 0.6 0.05 = 1.15

Since P ( A or B) > 1 is not possible, hence the given probabilities are not consistent.(b) P ( A or B ) = P ( A ) + P ( B ) P ( A and B )

= 0.5 + 0.4 0.1 = 0.8

which is less than 1.As the number of outcomes favourable to event 'A and B' should always be less than or equalto those favourable to the event A,

Therefore, P(AandB) P(A)

and similarly P(AandB) P(B)

In this case, P (A and B) = 0.1, which is less than both P (A) = 0.5 and P (B) = 0.4. Hence, theassigned probabilities are consistent.(c) In this case, P ( A and B) = 0.4, which is more than P (A ) = 0.2.

[ P(AandB) P(A)]∵

Hence, the assigned probabilities are not consistent.

Example 31.28 An urn contains 8 white balls and 2 green balls. A sample of three balls is

selected at random. What is the probability that the sample contains at least one green ball ?Solution : Urn contains 8 white balls and 2 green balls.

Total number of balls in the urn = 10Three balls can be drawn in 10C3 ways = 120 ways.

Let A be the event " at least one green ball is selected".Let us determine the number of different outcomes in A. These outcomes contain either onegreen ball or two green balls.

There are 2C1 ways to select a green ball from 2 green balls and for this remaining two white

balls can be selected in 8C2 ways.

Hence, the number of outcomes favourable to one green ball2 8

1 2C C

= 2 × 28 = 56

Similarly, the number of outcomes favourable to two green balls2 8

2 1C C 1 8 = 8

MATHEMATICS 545

Notes


Probability

Hence, the probability of at least one green ball isP ( at least one green ball )

= P ( one green ball ) + P ( two green balls )

=56 8

120 120

=64

120=

815

Example 31.29 Two balls are drawn at random with replacement from a bag containing 5

blue and 10 red balls. Find the probability that both the balls are either blue or red.

Solution : Let the event A consists of getting both blue balls and the event B is getting both redballs. Evidently A and B are mutually exclusive events.By fundamental principle of counting, the number of outcomes favourable to A 5 5 25.

Similarly, the number of outcomes favourable to B = 10 × 10 = 100.

Total number of possible outcomes = 15 × 15 = 225.

P (A) = 25 1

225 9 and P (B ) =

100 4225 9

.

Since the events A and B are mutually exclusive, therefore P ( A or B ) = P ( A ) + P ( B )

1 49 9

=59

Thus, P ( both blue or both red balls ) = 59

CHECK YOUR PROGRESS 31.31. A card is drawn from a well-shuffled pack of cards. Find the probability that it is a queen

or a card of heart.2. In a single throw of two dice, find the probability of a total of 7 or 12.3. The odds in favour of winning of Indian cricket team in 2010 world cup are 9 to 7. What

is the probability that Indian team wins ?4. The odds against the team A winning the league match are 5 to 7. What is the probability

that the team A wins the league match.5. Two dice are thrown. Getting two numbers whose sum is divisible by 4 or 5 is consid-

ered a success. Find the probability of success.6. Two cards are drawn at random from a well-shuffled deck of 52 cards with replacement.

What is the probability that both the cards are either black or red ?

MATHEMATICS

Notes


546

Probability

7. A card is drawn at random from a well-shuffled deck of 52 cards. Find the probabilitythat the card is an ace or a black card.

8. Two dice are thrown once. Find the probability of getting a multiple of 3 on the first dieor a total of 8.

9. (a) In a single throw of two dice, find the probability of a total of 5 or 7.(b) A and B are two mutually exclusive events such that P (A ) = 0.3 and P ( B) = 0.4.

Calculate P ( A or B ).10. A box contains 12 light bulbs of which 5 are defective. All the bulbs look alike and have

equal probability of being chosen. Three bulbs are picked up at random. What is theprobability that at least 2 are defective ?

11. Two dice are rolled once. Find the probability(a) that the numbers on the two dice are different,(b) that the total is at least 3.

12. A couple have three children. What is the probability that among the children, there willbe at least one boy or at least one girl ?

13. Find the odds in favour and against each event for the given probability

(a) P (A) = .7 (b) P (A) = 45

14. Determine the probability of A for the given odds(a) 7 to 2 in favour of A (b ) 10 to 7 against A.

15. If two dice are thrown, what is the probability that the sum is

(a) greater than 4 and less than 9 ?

(b) neither 5 nor 8 ?

16. Which of the following probability assignments are inconsistent ? Give reasons.(a) P (A) = 0.5 , P (B) = 0.3 , P (A and B ) = 0.4

(b) P (A) = P (B) = 0.4, P (A and B) = 0.2

(c) P (A) = 0.85, P (B) = 0.8, P (A and B) = 0.61

17. Two balls are drawn at random from a bag containing 5 white and 10 green balls. Findthe probability that the sample contains at least one white ball.

18. Two cards are drawn at random from a well-shuffled deck of 52 cards with replacement.What is the probability that both cards are of the same suit ?

31.6 SOME RESULTS ON PROBABILITY OF EVENTSResult 1 : Probability of impossible event is zero.Solution : Impossible event contains no sample points. Therefore, the certain event S and theimpossible event are mutually exclusive.

Hence, S S∪

P S P (S)∪

MATHEMATICS 547

Notes


Probability

or P S P ( ) P (S)P ( ) 0.

Result 2 : Probability of the complementary event A of A is given by

P A 1 P ( A )

Solution : A and A are disjoint events. Also,

A A S∪ P A A P ( S )∪

Using additive laws (ii) and (iii) , we get

P(A) P(A) 1 P(A) 1 P(A).

Result 3 : Prove that 0 P ( A ) 1, for any A in S.

Solution : We know that

A S P ( A ) P ( S )

P ( A ) 1 . [Using additive law]

We know that P ( A ) 0.

Hence, 0 P(A) 1.

Result 4 : If A and B are any two events, then

P A B P ( A ) P ( B ) P ( A B).∪ ∩

Solution :

Fig. 31.6

From the above figure, we can write

A B A A B∪ ∪ ∩

P (A B) P A A B .∪ ∪ ∩ ....(1)

Since the events A and (A B)∩ are disjoint, therefore law (iii) gives

P A A B P(A) P A B∪ ∩ ∩

Substituting this value in (1) , we get

P(A B) P(A) P A B∪ ∩

or P(A B) P(A) [P A B P(A B)] P(A B)∪ ∩ ∩ ∩ .....(2)

MATHEMATICS

Notes


548

Probability

From Fig. 31.6, we see that

(A B) A B B∩ ∪ ∩

P[(A B) A B ] P (B)∩ ∪ ∩ .

Further, the events (A B)∩ and A B∩ are disjoint, so from additive law, we get

P (A B) P A B P (B)∩ ∩

Substituting this value in (2), we get

P ( A B) P(A) P(B) P ( A B∪ ∩

Result 5 : If A and B are mutually exclusive events, then

P ( A B) P(A) P(B)∪

Solution : From additive law, we have

P ( A B) P(A) P(B) P(A B)∪ ∩ .....(1)

Since A and B are mutually exclusive events,

Therefore A B∩

P ( A B) P ( ) 0∩

Substituting this value in equation (1), we get

P ( A B) P(A) P(B)∪ ,

which is additive law for mutually exclusive events.

31.7 MULTIPLICATION LAW OF PROBABILITY FORINDEPENDENT EVENTS

Let us recall the definition of independent events.Two events A and B are said to be independent, if the occurrence or non-occurrence of onedoes not affect the probability of the occurrence (and hence non-occurrence) of the other.Can you think of some examples of independent events ?The event of getting 'H' on first coin and the event of getting 'T' on the second coin in a simultaneoustoss of two coins are independent events.What about the event of getting 'H' on the first toss and event of getting 'T' on the second tossin two successive tosses of a coin ? They are also independent events.Let us consider the event of 'drawing an ace' and the event of 'drawing a king' in two successivedraws of a card from a well-shuffled deck of cards without replacement.Are these independent events ?No, these are not independent events, because we draw an ace in the first draw with probability

452

. Now, we do not replace the card and draw a king from the remaining 51 cards and this

affect the probability of getting a king in the second draw, i.e., the probability of getting a king in

MATHEMATICS 549

Notes


Probability

the second draw without replacement will be 4

51.

Note : If the cards are drawn with replacement, then the two events become independent.Is there any rule by which we can say that the events are independent ?How to find the probability of simultaneous occurrence of two independent events?If A and B are independent events, then

P (A and B) P(A) . P (B)

or

P ( A B) P(A). P (B)∩

Thus, the probability of simultaneous occurrence of two independent events is the productof their separate probabilities.

Note : The above law can be extended to more than two independent events, i.e.,

P(A B C...) P(A) P(B) P(C)...∩ ∩

On the other hand, if the probability of the event 'A' and 'B' is equal to the product of theprobabilities of the events A and B, then we say that the events A and B are independent.

Example 31.30 A die is tossed twice. Find the probability of a number greater than 4 on

each throw.Solution : Let us denote by A, the event 'a number greater than 4' on first throw. B be the event'a number greater than 4' in the second throw. Clearly A and B are independent events.In the first throw, there are two outcomes, namely, 5 and 6 favourable to the event A.

2 1P(A)6 3

Similarly,1P(B)3

Hence, P ( A and B) P ( A ) . P ( B )

1 1.3 3

= 19

.

Example 31.31 Arun and Tarun appear for an interview for two vacancies. The probability

of Arun's selection is 13

and that of Tarun's selection is 15

. Find the probability that

(a) both of them will be selected.(b) none of them is selected.(c) at least one of them is selected.(d) only one of them is selected.

Solution : Probability of Arun's selection 1P(A)3

MATHEMATICS

Notes


550

Probability

Probability of Tarun's selection 1P(T)5

(a) P (both of them will be selected ) = P (A) P (T)1 13 5

= 1

15(b) P ( none of them is selected )

P A P T

1 11 1

3 5

2 43 5

815

(c) P (at least one of them is selected )1 P ( None of them is selected )

1 P A P T

1 11 1 1

3 5

2 41

3 5

8115

715

(d) P ( only one of of them is selected )

P A P T P A P(T)

1 4 2 13 5 3 56 2

15 5

Example 31.32 A problem in statistics is given to three students, whose chances of solving it

are1 1,2 3

and 14

respectively. What is the probability that problem will be solved ?

Solution : Let 1 2p , p and 3p be the probabilities of three persons of solving the problem.

Here, 11p2

, 21p3

and 31p4

.

The problem will be solved, if at least one of them solves the problem.

MATHEMATICS 551

Notes


Probability

P ( at least one of them solves the problem )

1 P (None of them solves the problem) .....(1)

Now, the probability that none of them solves the problem will be

P ( none of them solves the problem ) 1 2 31 p 1 p 1 p

1 2 32 3 4

14

Putting this value in (1), we get

P ( at least one of them solves the problem )114

34

Hence, the probability that the problem will be solved is 34

.

Example 31.33 Two balls are drawn at random with replacement from a box containing 15

red and 10 white balls. Calculate the probability that(a) both balls are red.(b) first ball is red and the second is white.(c) one of them is white and the other is red.Solution :(a) Let A be the event that first drawn ball is red and B be the event that the second ball drawnis red. Then as the balls drawn are with replacement,

therefore15 3 3P(A) ,P(B)25 5 5

As A and B are independent eventstherefore P ( both red ) = P ( A and B )

P(A) P(B)

3 3 95 5 25

(b) Let A : First ball drawn is red.B : Second ball drawn is white.

P ( A and B ) P(A) P(B)

3 2 65 5 25

.

(c) If WR denotes the event of getting a white ball in the first draw and a red ball in the seconddraw and the event RW of getting a red ball in the first draw and a white ball in the second draw.Then as 'RW' and WR' are mutually exclusive events, therefore

MATHEMATICS

Notes


552

Probability

P ( a white and a red ball )= P ( WR or RW )= P (WR) + P (RW)= P (W) P (R) + P (R) P (W)

2 3 3 2. .5 5 5 5

=6 625 25

1225

Example 31.34 The odds against Manager X settling the wage dispute with the workers are8 : 6 and odds in favour of manager Y settling the same dispute are 14 : 16.(i) What is the chance that neither settles the dispute, if they both try independently of each

other ?(ii) What is the probability that the dispute will be settled ?Solution : Let A be the event that the manager X will settle the dispute and B be the event thatthe manager Y will settle the dispute. Then, clearly

(i)6 3 3 4P(A) , P A 1 P(A) 1

14 7 7 7

14 7 14 16 8P(B) , P B 130 15 30 30 15

The required probability that neither settles the dispute is given by

P(A B) P A P B∩

4 8.7 15

32105

(Since A, B are independent, therefore, A , B also independent)

(ii) The dispute will be settled, if at least one of the managers X and Y settles the dispute.Hence, the required probability is given by

P (A B)∪ = P [At least one of X and Y settles the dispute]

=1 P [None settles the dispute]

1 P A B∩

321105

73105

Example 31.35 A dice is thrown 3 times. Getting a number '5 or 6' is a success. Find the

probability of getting(a) 3 successes (b) exactly 2 successes (c) at most 2 successes (d) at least 2 successes.Solution : Let S denote the success in a trial and F denote the ' not success' i.e. failure.Therefore,

MATHEMATICS 553

Notes


Probability

2 1P ( S )6 3

1 2P ( F ) 13 3

(a) As the trials are independent, by multiplication theorem for independent events,P ( SSS ) = P ( S ) P (S ) P ( S )

1 1 1 13 3 3 27

P ( SSF ) = P ( S ) P ( S) P ( F )1 1 23 3 3

227

Since the two successes can occur in 32C ways

P (exactly two successes) 32

2C27

29

(c) P ( at most two successes) 1 P(3successes)

1127

=2627

(d) P ( at least two successes) = P ( exactly 2 successes) + P ( 3 successes)2 19 27

727

Example 31.36 A card is drawn from a pack of 52 cards so that each card is equally likely

to be selected. Which of the following events are independent ?(i) A : the card drawn is a spade

B : the card drawn is an ace(ii) A : the card drawn is black

B : the card drawn is a king(iii) A : the card drawn is a king or a queen

B : the card drawn is a queen or a jackSolution : (i) There are 13 cards of spade in a pack.

13 1P (A)52 4

There are four aces in the pack.

4 1P (B)52 13

A B∩ { an ace of spade }

MATHEMATICS

Notes


554

Probability

1P (A B)52

∩

Now1 1 1P (A) P ( B )4 13 52

Since P (A B) P(A). P (B)∩

Hence, the events A and B are independent.(ii) There are 26 black cards in a pack.

26 1P (A)52 2

There are four kings in the pack.4 1P (B)

52 13A B∩ = { 2 black kings }

2 1P (A B)52 26

∩

Now,1 1 1P ( A ) P ( B )2 13 26

Since P (A B) P ( A ) P ( B )∩

Hence, the events A and B are independent.(iii) There are 4 kings and 4 queens in a pack of cards.

Total number of outcomes favourable to the event A is 8.8 2P ( A )

52 13

Similarly,2P ( B )

13A B∩ { 4 queens }

4 1P(A B)52 13

∩

2 2 4P ( A ) P ( B )13 13 169

Here, P (A B) P ( A ) . P ( B )∩Hence, the events A and B are not independent.

Example 31.37 Consider a group of 36 students. Suppose that A and B are two propertiesthat each student either has or does not have. The events are

A : Student has blue eyesB : Student is a male

MATHEMATICS 555

Notes


Probability

Out of 36, there are 12 male and 24 female students and half of them in each has blue eyes. Arethese events independent ?Solution : With regard to the given two properties, i.e., either has or does not have, the 36students are distributed as follows :

Blue eyes Not blue eyes Total

A (A)

Male (B) 6 6 12

Female (B) 12 12 24

Total 18 18 36

If we choose a student at random, the probabilities corresponding to the events A and B are18 1P ( A )36 212 1P ( B )36 3

6 1P (A B)36 6

∩

Also1 1 1P(A) P(B)2 3 6

Here, P(A B) P (A) P(B)∩

Hence, the events A and B are independent.

Example 31.38 Suppose that we toss a coin three times and record the sequence of heads

and tails. Let A be the event ' at most one head occurs' and B the event ' both heads and tailsoccur'. Are these event independent ?

Solution : The sample space in tossing a coin three times will be

S = {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT}Also, A B∩ = { TTH, THT, HTT }

4 1P (A)8 2

,6 3P (B)8 4

,3P (A B)8

∩

Moreover,1 3P (A) P ( B )2 4

38

Which equals P (A B)∩ . Hence, A and B are independent.

CHECK YOUR PROGRESS 31.41. A husband and wife appear in an interview for two vacancies in the same department.

MATHEMATICS

Notes


556

Probability

The probability of husband's selection is 17

and that of wife's selection is 15

. What is the

probability that(a) Only one of them will be selected ?(b) Both of them will be selected ?(c) None of them will be selected ?(d) At least one of them will be selected ?

2. Probabilities of solving a specific problem independently by Raju and Soma are 12

and

13

respectively. If both try to solve the problem independently, find the probability that

(a) the problem is solved.(b) exactly one of them solves the problem.

3. A die is rolled twice. Find the probability of a number greater than 3 on each throw.4. Sita appears in the interview for two posts A and B, selection for which are independent.

The probability of her selection for post A is 15

and for post B is 17

. Find the probability

that she is selected for(a) both the posts(b) at least one of the posts.

5. The probabilities of A, B and C solving a problem are 1 2,3 7

and 38

respectively. If all

the three try to solve the problem simultaneously, find the probability that exactly one ofthem will solve it.

6. A draws two cards with replacement from a well-shuffled deck of cards and at the sametime B throws a pair of dice. What is the probability that(a) A gets both cards of the same suit and B gets a total of 6 ?(b) A gets two jacks and B gets a doublet ?

7. Suppose it is 9 to 7 against a person A who is now 35 years of age living till he is 65 and3:2 against a person B now 45 living till he is 75. Find the chance that at least one of thesepersons will be alive 30 years hence.

8. A bag contains 13 balls numbered from 1 to 13. Suppose an even number is considereda 'success'. Two balls are drawn with replacement, from the bag. Find the probability ofgetting(a) Two successes (b) exactly one success(c) at least one success (d) no success

9. One card is drawn from a well-shuffled deck of 52 cards so that each card is equallylikely to be selected. Which of the following events are independent ?(a) A : The drawn card is red B : The drawn card is a queen

(b) A : The drawn card is a heart B: The drawn card is a face card

MATHEMATICS 557

Notes


Probability

31.8 CONDITIONAL PROBABILITYSuppose that a fair die is thrown and the score noted. Let A be the event, the score is 'even'.Then

A = {2 , 4 , 6 }3 1P (A)6 2

.

Now suppose we are told that the score is greater than 3. With this additional information whatwill be P (A ) ?Let B be the event, 'the score is greater than 3'. Then B is {4, 5, 6 }. When we say that B hasoccurred, the event 'the score is less than or equal to 3' is no longer possible. Hence the samplespace has changed from 6 to 3 points only. Out of these three points 4, 5 and 6; 4 and 6 areeven scores.

Thus, given that B has occurred, P (A) must be 23

.

Let us denote the probability of A given that B has already occurred by P ( A | B) .

FIg. 31.7

Again, consider the experiment of drawing a single card from a deck of 52 cards. We areinterested in the event A consisting of the outcome that a black ace is drawn.Since we may assume that there are 52 equally likely possible outcomes and there are twoblack aces in the deck, so we have

2P (A)52

.

However, suppose a card is drawn and we are informed that it is a spade. How should thisinformation be used to reappraise the likelihood of the event A ?Clearly, since the event B "A spade has been drawn " has occurred, the event "not spade" is nolonger possible. Hence, the sample space has changed from 52 playing cards to 13 spadecards. The number of black aces that can be drawn has now been reduced to 1.Therefore, we must compute the probability of event A relative to the new sample space B.Let us analyze the situation more carefully.The event A is " a black ace is drawn'. We have computed the probability of the event Aknowing that B has occurred. This means that we are computing a probability relative to a newsample space B. That is, B is treated as the universal set. We should consider only that part ofA which is included in B.

MATHEMATICS

Notes


558

Probability

Fig. 31.8

Hence, we consider A B∩ (see figure 31.8).Thus, the probability of A. given B, is the ratio of the number of entries in A B∩ to thenumber of entries in B. Since n ( A B)∩ 1 and n ( B ) 13 ,

thenn(A B)P ( A | B )

n(B)∩ 1

13

Notice that n A B 1∩1P ( A B )

52∩

n ( B ) 1313P ( B )52

11 P(A B)52P (A | B) 1313 P(B)

52

∩.

This leads to the definition of conditional probability as given below :Let A an B be two events defined on a sample space S. Let P(B) > 0, then the conditionalprobability of A, provided B has already occurred, is denoted by P(A|B) and mathematicallywritten as :

P (A | B) =P(A B)

P(B)∩

, P (B) 0

Similarly,P (A B)P ( B | A ) , P(A) 0

P(A)∩

The symbol P( A | B ) is usually read as "the probability of A given B".

Example 31.39 Consider all families "with two children (not twins). Assume that all theelements of the sample space {BB, BG, GB,GG} are equally likely. (Here, for instance, BGdenotes the birth sequence "boy girls"). Let A be the event {BB} and B be the event that 'atleast one boy'. Calculate P (A | B ).

Solution : Here, A = { BB }B = { BB, BG, GB }

A B {BB}∩

1P (A B)4

∩

MATHEMATICS 559

Notes


Probability

1 1 1 3P(B)4 4 4 4

Hence,P(A B)P ( A | B )

P(B)∩

1434

13

.

Example 31.40 Assume that a certain school contains equal number of female and malestudents. 5 % of the male population is football players. Find the probability that a randomlyselected student is a football player male.Solution : Let M = Male

F = Football playerWe wish to calculate P (M F)∩ . From the given data,

1P (M)2

( ∵ School contains equal number of male and female students)

P ( F|M ) = 0.05But from definition of conditional probability, we have

P M FP F | M

P M∩

P M F P M P F | M∩

1 0.05 0.0252

Example 31.41 If A and B are two events, such that P (A) = 0.8,

P ( B ) = 0.6, P ( A B) 0.5∩ , find the value of

(i) P (A B)∪ (ii) P ( B | A ) (iii) P ( A | B ).

Solution :(i) P A B P A P B P A B∪ ∩

0.8 0.6 0.5 0.9

(ii)P A B

P B | AP A∩ 0.5 5

0.8 8

(iii)P A B

P A | BP B∩

0.50.6

56

MATHEMATICS

Notes


560

Probability

Example 31.42 Find the chance of drawing 2 white balls in succession from a bag containing5 red and 7 white balls, the balls drawn not being replaced.Solution : Let A be the event that ball drawn is white in the first draw. B be the event that balldrawn is white in the second draw.

P A B P A P B | A∩

Here,7 6P A , P B | A

12 117 6 7P A B

12 11 22∩

Example 31.43 A coin is tossed until a head appears or until it has been tossed three times.Given that head does not occur on the first toss, what is the probability that coin is tossed threetimes ?Solution : Here, it is given that head does not occur on the first toss. That is, we may get thehead on the second toss or on the third toss or even no head.Let B be the event, " no heads on first toss".

Then B = {TH, TTH, TTT }These events are mutually exclusive.

P B P TH P TTH P TTT .....(1)

Now1P TH4

(∵ This event has the sample space of four outcomes)

and1P TTH P TTT8

(∵ This event has the sample space of eight outcomes)

Putting these values in (1), we get

1 1 1 4 1P B4 8 8 8 2

Let A be the event "coin is tossed three times".

Then A TTH,TTT

We have to find P (A | B) .

P A BP A | B

P B∩

Here, A B∩ = A

114P A | B 1 2

2

MATHEMATICS 561

Notes


Probability

CHECK YOUR PROGRESS 31.51. A sequence of two cards is drawn at random (without replacement) from a well-shuffled

deck of 52 cards. What is the probability that the first card is red and the second card isblack ?

2. Consider a three child family for which the sample space is{ BBB, BBG, BGB, GBB, BGG, GBG, GGB, GGG }

Let A be the event " the family has exactly 2 boys " and B be the event " the first child isa boy". What is the probability that the family has 2 boys, given that first child is a boy ?

3. Two cards are drawn at random without replacement from a deck of 52 cards. What isthe probability that the first card is a diamond and the second card is red ?

4. If A and B are events with P (A) = 0.4, P (B) = 0.2, P ( A B) 0.1∩ , find the prob-ability of A given B. Also find P ( B|A ).

5. From a box containing 4 white balls, 3 yellow balls and 1 green ball, two balls are drawnone at a time without replacement. Find the probability that one white and one yellow ballis drawn.

31.9 THEOREMS ON MULTIPLICATION LAW OFPROBABILITY AND CONDITIONAL PROBABILITY.

Theorem 1 : For two events A and B,P A B P A .P B | A ,∩

and P A B P B .P A | B ,∩

where P ( B|A ) represents the conditional probability of occurrence of B, when the event A hasalready occurred and P ( A|B ) is the conditional probability of happening of A, given that B hasalready happened.Proof : Let n (S) denote the total number of equally likely cases, n (A) denote the casesfavourable to the event A, n (B) denote the cases favourable to B and n ( A B)∩ denote thecases favourable to both A and B.

n AP A

n S ,

n BP B

n S

n A BP A B

n S∩

∩ .....(1)

For the conditional event A|B , the favourable outcomes must be one of the sample points of B,i.e., for the event A|B, the sample space is B and out of the n (B) sample points, n ( A B)∩pertain to the occurrence of the event A, Hence,

MATHEMATICS

Notes


562

Probability

n A BP A | B

n B∩

Rewriting (1), we getn B n A B

P A B P(B). P(AB)n S n B

∩∩

Similarly, we can proveP A B P A .P B | A∩

Note : If A and B are independent events, thenP A | B P A and P B | A P B

P A B P A .P B∩Theorem 2 : Two events A and B of the sample space S are independent, if and only if

P A B P A .P B∩Proof : If A and B are independent events,then P A | B P A

We know thatP A B

P A | BP B∩

P A B P A P B∩Hence, if A and B are independent events, then the probability of 'A and B' is equal to theproduct of the probability of A and probability of B.Conversely, if P A B P A .P B∩ , then

P A BP A | B

P B∩

gives

P A P BP A | B

P B= P (A)

That is, A and B are independent events.

LET US SUM UP

Events Relation : The complement of an event A consists of all those outcomes whichare not favourable to the event A, and is denoted by 'not A' or by A .Event 'A or B' : The event 'A or B' occurs if either A or B or both occur.Event 'A and B' : The event 'A and B' consists of all those outcomes which are favourableto both the events A and B.Addition Law of Probability : For any two events A and B of a sample space S

P ( A or B ) = P (A) + P ( B ) P ( A and B)Additive Law of Probability for Mutually Exclusive Events : If A and B are twomutually exclusive events, then

LET US SUM UP

MATHEMATICS 563

Notes


Probability

P ( A or B ) = P (A B)∪ = P ( A ) + P ( B ).Odds in Favour of an Event : If the odds for A are a to b, then

aP(A)a b

If odds against A are a to b, then

bP(A)a b

Two events are mutually exclusive, if occurrence of one precludes the possibility ofsimultaneous occurrence of the other.Two events A and B are said to be independent, if the occurrence or non-occurrence ofone does not affect the probability of the occurrence (and hence non-occurrence) of theother.If A and B are independent events, then

P ( A and B ) = P ( A ). P ( B )

or P(A B) P ( A ) . P ( B )∩

For two events A and B,

P(A B) P ( A ) P ( B | A )∩ , P (A) 0

or P(A B) P ( B ) P ( A | B)∩ , P (B) 0

where P (B|A) represents the conditional probability of occurrence of B, when the eventA has already happened and P (A|B) represents the conditional probability of happeningof A, given that B has already happened.


TERMINAL EXERCISE

1. In a simultaneous toss of four coins, what is the probability of getting(a) exactly three heads ? (b) at least three heads ?(c) atmost three heads ?

2. Two dice are thrown once. Find the probability of getting an odd number on the first dieor a sum of seven.

3. An integer is chosen at random from first two hundred integers. What is the probabilitythat the integer chosen is divisible by 6 or 8 ?

4. A bag contains 13 balls numbered from 1 to 13. A ball is drawn at random. What is the


MATHEMATICS

Notes


564

Probability

probability that the number obtained it is divisible by either 2 or 3 ?5. Find the probability of getting 2 or 3 heads, when a coin is tossed four times.6. Are the following probability assignments consistent ? Justify your answer.

(a) P (A) = 0.6, P (B) = 0.5 , P ( A and B ) = 0.4(b) P (A) = 0.2, P (B) = 0.3 , P (A and B ) = 0.4(c) P (A) = P (B) = 0.7, P ( A and B ) = 0.2

7. A box contains 25 tickets numbered 1 to 25. Two tickets are drawn at random. What isthe probability that the product of the numbers is even ?

8. A drawer contains 50 bolts and 150 nuts. Half of the bolts and half of the nuts are rusted.If one item is chosen at random, what is the probability that it is rusted or is a bolt ?

9. A lady buys a dozen eggs, of which two turn out to be bad. She chose four eggs toscramble for breakfast. Find the chances that she chooses(a) all good eggs (b) three good and one bad eggs(c) two good and two bad eggs (d) at least one bad egg.

10. Two cards are drawn at random without replacement from a well-shuffled deck of 52cards. Find the probability that the cards are both red or both kings.

11. Three groups of children contain respectively 3 girls and 1 boy, 2 girls and 2 boys, 1 girland 3 boys. One child is selected at random from each group. Show the chances that

three selected children consist of 1 girl and 2 boys is 1332

.

12. A die is thrown twice. Find the probability of a prime number on each throw.13. Kamal and Monika appears for an interview for two vacancies. The probability of Kamal's

selection is 13

and that of Monika's rejection is 45

. Find the probability that only one of

them will be selected.14. A bag contains 7 white, 5 black and 8 red balls. Four balls are drawn without replacement.

Find the probability that all the balls are black.15. For two events A and B, it is given that

P (A) = 0.4, P (B) = p and P (A B)∪ = 0.6

(a) Find p so that A and B are independent events.(b) For what value of p, of A and B are mutually exclusive ?

16. The odds against A speaking the truth are 3 : 2 and the odds against B speaking the truthare 5 : 3. In what percentage of cases are they likely to contradict each other on aidentical issue ?

17. Let A and B be the events such that1 2P A , P B ,

2 31P A B4

∩ .

Compute P ( A|B) and P ( B|A ).

MATHEMATICS 565

Notes


Probability

ANSWERS


1. 16

2.12

3. 12

4.34

5. (i)35

(ii)25

6. (i)5

36(ii)

536

(iii)1

12(iv)

136

7.59

8.1

129.

12

10. (i)14

(ii)1

13(iii)

152

11. (i)5

12(ii)

16

(iii)1136

12. (i)18

(ii)78

(iii)18


1.18

2.2039

3.(a)4

25(b)

38245

4.1

5525 6. (i)

310

(ii) 16

(iii)1

30

7.10

1338.

47

9.60

14310.

14


1.4

132.

736

3.9

164.

712

5.49

6.12

7.7

138.

512

9. (a)5

18(b) 0.7 10.

411

11. (a)56

(b)3536

12.34

13. (a) The odds for A are 7 to 3 . The odds against A are 3 to 7

MATHEMATICS

Notes


566

Probability

13. (b) The odds for A are 4 to 1 and The odds against A are 1 to 4

14. (a)79

(b) 7

1715. (a)

59

(b)34

16. (a), (c) 17.47

18.14


1. (a)27

(b)1

35(c)

2435

(d)1135

2. (a)23

(b)12

3 .14

4. (a)1

35(b)

1135

5.12

6. (a)5

144(b)

11014

7.5380

8. (a)36

169(b)

84169

(c)120169

(d)49

1699. (a) Independent (b) Independent


1.1351

2.12

3.25

204

4.12

,14

5.37

TERMINAL EXERCISE

1. (a)14

(b)5

16(c)

1516

2.7

12

3.14

4.8

135.

58

6. Only (a) is consistent

7.456625

8.58

9. (a)1433

(b)1633

(c)1

11(d)

1933

10.55221

12.14

13.2

1514.

1969

15.13

16.1940

17.34

,12

respectively..

567

Random Variables andProbability Distributions

31.10 Definition

Let S be the sample space associated with a random experiment A function X : S → R is called

a random variable.

Note : If X is a random variable then X−1(P(R)) = P(S)

Here P stands for the probability function and P(S) stands for the power set of S .

Example 1

Let S be the sample space of the experiment of rolling a fair die.

Then X : S → R given by X(n) = 0, if n is even.

= 1, If n X : S → is odd.

is a random variable.

Here S = {1, 2, 3, 4, 5, 6} and X(1) = X(3) = X(5) = 1; X(2) = X(4) = X(6) = 0.

31.10.1 Definition

Suppose X : S → R is a random variable. If the range of X is either finite or countably infinite,

then X is called a discrete random variable.

A random variable which can take all real values in an interval (a, b ) is called a conitnuous

random variable.

568

31.10.2 Definition

The mean (µ ) and variance (σ2) of a discrete random variable X are

µ = (X )n nx p x=∑ , σ2 =2

( ) (X )n nx p x−µ =∑

σ2 = 2 2(X )n nx p x = −µ ∑The standard deviation σ is the square root of the variance.

Example 2

A cubical die is thrown. Find the mean and variance of X, giving the number on the face that

shows up.

Sol : Let S be the sample space and X be the random variable associated with S, where p(X) is given

by the following table

X = xi 1 2 3 4 5 6

P(X = xi)1

6

1

6

1

6

1

6

1

6

1

6

Mean of X =6

i = 1

X P(X )i ixµ = =∑

=1 1 1 1 1 1

1. 2. 3. 4. 5. 6.6 6 6 6 6 6

+ + + + +

=1 6 7

6 2

×

=7

2

Variance of X = σ2 = 6

2 2

i = 1

P(X )i ix x= −µ∑

22 2 2 2 2 21 1 1 1 1 1 7

1 . 2 . 3 . 4 . 5. 6 .6 6 6 6 6 6 2

= + + + + + −

=1 6 7 13 49 35

6 6 4 12

× × − =

569

31.11 Theoretical discrete distributions Binomial and possion distributions

Suppose that an experiment has only two possible outcomes. For instance, when a coin is

tossed the possible out comes are head and tail. Each performance of an experiment with two possible

outcomes are termed as success and failure. If p is taken as the probability of success and q is the

probability of failure, it follows that p + q = 1. Many problems can be solved by determining the

probability of x successes when an experiment consists of n independent Bernoulli trials.

We shall discuss two theoretical frequency distributions in this section. One is Binomial distribu-

tion and the other is poission distribution.

31.11.1 Binomial distribution

Definition

A discrete random variable X is said to follow a binomial distribution (or simply a binomial

variable with parameters n and p) where 0 < p < 1 if

P(X = x) = nCxpx qn−x, x ∈ {0, 1, 2, ..., n}

If X is a binomial variate with parameters n, p; then it is also described by writing X ~ B (n, p)

Theorem

If X ~ B (n, p), then the mean µ and the variance σ2 of X are equal to np and npq

respectively.

n is number of trials, p be the probability of sucess and q be the probability of failure.

31.12 Poisson distribution

In binomial distribution, the number of trials n is precisely known and is finite. But there are some

situations where this may not be possible. Also if the event is rare and casual, the successful events are

quite a few in the sample space. If we know the number of times of times an event occurred but not

how many times it did not occur, then the bonomial distribution is inapplicable. In such cases we

compute the binomial probabilities approximately by using Poisson distribution

570

Objectives

From this chapter we can get to know following points

� the number of telephone calls received at a telephone exchange in a given time interval.

� the number of printing errors in a page of a book.

� the number of road/rail accidents reported in a city in a given period of time (say a day)

� The number of defective blades in a packet of 100.

31.12.1 Definition

If X is a discrete random variable that can assume value 0, 1, 2, 3, ..... such that for some fixedλ > 0.

p(X = x) = , 0, 1, 2.......xe

xx

−λ λ = then X is said to follow a Possion distribution with param-

eter λ then its mean µ, variance σ2 = λ.

Example 3

8 coins are tossed simultaneously. If the number of heads turned up is denoted by the variable X.

Then find the mean and variance of X.

Sol : Here n = 8

prab. of geting in a head on a coin 1

2p =

1 11 , 8

2 2p q p n= ⇒ = − = =

Mean (µ) np = 8 1

42

=

Variance (σ2) = npq = 8 1 1

22 2

=

571

Example 4

Find the maximum number of times a fair coin must be tossed so that the probability of geting

atleast one head is atleast 0.8.

Sol : Let n be the number of times a fair coin tossed X denotes the number of heads getting X follows

binomial distribution with parameters n and

1 11 , (X 1) 0.8

2 2p q p p= ⇒ = − = ≥ ≥

⇒ 1 − p(X = 0) > 0.8

⇒ p(X = 0) < 0.2

0C1

0.22

n

n ⇒ ≤

1 1

52n⇒ ≤

15

2n⇒ ≥

∴ The Maximum value of n is 3.

Example 5

Find the parameters of the binomial variate whose mean and variance are 15 15

,2 4

respectively.

Sol : n, p are parameters of the binomial distribution = n p=15

2

its mean (µ) = npq = 15

4

1514

15 22

npq

np

= =

572

1 1 1, 1 1

2 2 2q p q⇒ = = − = − =�

∴ n p=15

2

1 1515

2 2n n

⇒ = ⇒ =

∴ n = 15, p = 1

2.

Example 6

If X is a binomial variate with 9 p(X = 4) = p(X = 2) and n = 6 then, find the parameter p.

Sol : Given p + q = 1

� 9 (X = 4) = p(X = 2)

⇒ 9(6C4

) q2 . p4 = 6C2q4 p2

24 2

4 29

q p q

pp q

⇒ = =

3 1 3q

p pp

⇒ = ⇒ − =

1

4p∴ =

3

4q⇒ =

∴ Parameter1

4p =

Example 7

If X is a poisson variate such that p(X = 0) = p(X = 1) = k, Then show that k = e−1.

Sol : Let λ = 0 be the parameter of a poisson variate X

573

Given p(X = 0) = p(X = 1) = k

1. .

0! 1!e e

k−λ 0 −λλ λ= =

⇒ e−λ = e−λ . λ ⇒ λ = 1

11

0!e

k e− 0

−λ∴ = =

1k e−∴ = .

Example 8.

If X follows a poisson distribution and p(X = 1) = 3 p(X = 2), Then find the variance of X.

Sol : Let λ > 0 be the parameter of a poission variate X

Given p(X = 1) = 3(p(x = 2))

1 2. .3.

1! 2!e e−λ −λλ λ=

23

⇒ λ =

∴ Variance 2

3λ =

Check Your Progress

1. 8 coins are tossed simultaneously. Find the probability of getting atleast 6 heads.

2. The mean and variance of a binomial distribution are 4 and 3 respectively Fix the distribution and

find p(X > 1).

3. If X − B(n, p), µ = 20, σ2 = 10, then find n and p .

574

4. For a poisson variate X, p(X = 2) = p(X = 3). Find the variance of X.

5. A poisson variable satisfies p(X = 1) = p(X = 2) Find p(X = 5) .

Let Us Sum Up

1. If p is the probability of a success, q be the probability of a failure such the p + q = 1 and n is the

number of Bernoulli trials, then the probability distribution of a discrete random variable X, called

a binomial variate is given by

p(X = k) = nCk pk qn−k, k = 0, 1, 2, ..., n

This is called the binomial distribution.

Here n and p are called the parameters of the distribution.

In this case X is expressed as X~ B(n, p).

2. If X is a binomial variate with parameters n and p i.e., X∼ B(n, p), then the mean of the

distribution µ = npand the variance of the distribution σ2 = npq The standard deviation of this

distribution is given by npq .

3. The probability distribution of a discrete random variable X (called the Poisson variable) given

by p(X = k) = !

ke

k

−λ λ, k = 0, 1, 2, ....and λ = 0,is called the Poisson distribution.

Here λ is called the parameter of X .

4. If x is a Poisson variate with parameter λ then its mean µ = variance σ2 = λ .

Supportive Websites



575

Answers

Check Your Progress

(1)37256

(2)16

31

4 −

(3) n = 40, p = 1

2

(4) 3

(5)2 52

5 !e−

576

MODEL QUESTION PAPER

Intermediate (TOSS) MATHEMATICS 311 (E) (Set II)

Time: 3 hrs Max. Marks: 100

Instructions:-

1) All questions are Compulsory. 2) This question paper consists of two parts viz., A and B

3) All questions from Part A are to be attempted

4) Part-B has two options. Candidates are required to attempt questions from one option only.

(Part A)

SECTION 1

Each question carries 1 mark. (13 x 1 = 13 marks)

1) If Z1 = x + iy, and Z2 = a + ib then what is Z1 + Z2 ?

2) What are the roots of x2 + 5x + 6 = 0 ?

3) Show the region represented by the in-equations x 0 and y 0

4) t of their slopes?

5) Find the distance between the points (-1,2) and (0,-6) ?

6) Convert 2700 into Radians.

7) What is the domain of the function?

8) Evaluate Lt 0[ 2 + 1 3 5]

9) What is the derivative of w.r.t x ?

10)

formulae of slope of the tangent of the curve y = f (x) at (x1, y1)

11) What is + 1 ? . . ]

12) Write the Integrating factor of ..

13) What is the principle of standard deviation for grouped data?

577

Section 2 ( Each question carries 2 marks)

14) If A = 1 2

1 02x2 and B = 2 1

2 22x2 then prove that AB BA

15) What is the expanded form of 11 12 13

21 22 23

31 32 33

16) Find the inverse of the matrix

17) Find the equation of the parabola whose focus is (2,3), directrix is 3x + 4y = 1.

18)

19) Draw the Venn diagram for A B, when A and B are disjoint sets.

20) What is the principle value of cos 1(3

2)

21) If y = cos 1(4 3 3 ) find .

22) What is the Area bounded by the curve x = y, Y-axis and the line y = 0, y = 3.

23) A die is tossed twice. Find the probability of a number greater than 4 in each throw.

Section 3

(Each question carries 5 marks)

24) There are 5 Mathematics, 4 Physics and 5 Chemistry books. In how many ways can you arrange

4 Mathematics, 3 Physics and 4 Chemistry books, if the books on the same subjects are arranged

together?

25) If the 4th term and the 9th term of a G.P. are 8 and 256 respectively, then write the G.P.

26) Find the Maximum and Minimum value of the function f (x) = Sin x (1 + Cos x) in (0, )

27) The data below presents the earning of 50 labours of a factory. Calculate the Mean deviation.

Earning in Rs: 1200 1300 1400 1500 1600 1800

No. of Labors: 4 7 15 12 7 3

578

Section 4

(Each question carries 8 marks)

28) Using Principle of Mathematical induction, prove the following

1

1.3+

1

3.5+

1

5.7+ +

1

2 1 . 2 +1 =

n

2 +1

29) Find the equation of the circle which passes through the points (0,2), (2,0) and (0,0)

30) In , < = 600 prove that +

+c

+= 1

31) If y = ( ) + ( ) then find .

Part B

Optional I (VECTORS and 3 D Geometry)

32) If 1 = - + and 2 = 2 - 4 - 3 , then the magnitude of 1 + 2 1 Mark

33) Write the position vector of the centroid of the triangle, whose vertices are A ( ), B ( ), C ( ) 4

Marks.

34) If the foot of the perpendicular drawn from the origin to the plane is (4, -2, -5), then find the

equation of the plane. (5 Marks)

35) Find the centre and radius of the circle x2 + y2 + z2 6x 4y + 12z 36 = 0, x + 2y 2z = 1

(Optional II)

32) Define Share 1 Mark

33) Write any one use of Index numbers 1 Mark

34) Discuss the characteristics of VAT 5 Marks

35) For a new product a manufacturer spends Rupees 1,00,000 on the infrastructure and the

variable cost is estimated as Rs. 150 per unit of the product the sale price per unit fixed at Rs.

200.

Calculate (1) Cost function

(2) Revenue function

(3) Profit function and

(4) the breakeven point

intermediate (toss) course senior secondary course

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